GRAPHICS 


AND 


STRUCTURAL  DESIGN 


BY 

H.  D.  HESS,  M.E. 

Professor  of  Machine  Design,  Sibley  College,  Cornell  University.     Formerly  Designer 

and  Computer  for  the  Mechanical  Department  of  the  Pencoyd  Iron 

Works  and  the  American  Bridge  Co.     Member  of  the 

American  Society  of  Mechanical  Engineers. 


FIRST  EDITION 

FIRST    THOUSAND 


NEW  YORK 

JOHN   WILEY   &    SONS,  INC. 
LONDON:   CHAPMAN  &  HALL,  LIMITED 


COPYRIGHT,  1913, 

BY 
H.  D.  HESS 


Stanhope 

F.    H.  GILSON   COMPANY 
BOSTON,  U.S.A. 


PREFACE 


THIS  text  is  intended  for  the  author's  classes  in  General 
Engineering  Design  in  Sibley  College,  Cornell  University.  The 
treatment  of  the  subject  has  been  kept  as  general  as  possible  and 
is  that  used  for  his  students  during  the  past  four  or  five  years. 
The  problems  chosen  for  discussion  are  those  on  the  border  line 
between  Civil  and  Mechanical  Engineering.  To  the  general 
designer  knowledge  of  this  subject  is  indispensable  while  acquaint- 
ance with  the  methods  used  in  determining  the  stresses  in,  and 
the  subsequent  designing  of,  structures  is  of  the  greatest  benefit 
to  others  in  all  designing  for  strength  whether  of  machines  or 
structures. 

Although  the  book  is  called  " Graphics  and  Structural  Design" 
the  determination  of  the  stresses  has  not  been  confined  to  graphi- 
cal means;  the  other  usual  methods  have  been  included. 

The  design  of  a  plate  girder  may  seem  out  of  keeping  with 
the  purpose  of  the  book.  The  author  formerly  used  a  crane 
runway  girder  instead  of  the  railway  bridge,  but  decided  that 
the  railway  girders  made  the  more  comprehensive  and  better 
problem.  The  design  of  the  latter  permitted  the  use  of  the 
moment  table  and  acquainted  the  student  with  the  usual  method 
of  treating  locomotive  and  train  loads. 

Incidentally,  one  who  could  design  a  railway  plate  girder  was 
well  prepared  to  design  a  runway  girder,  although  the  reverse 
was  not  so  generally  true. 

The  author's  practice  is  to  use  a  set  of  problems  paralleling  the 
drawing-room  work,  to  assign  a  number  of  these  problems  each 
week  to  the  students  upon  which  they  recite.  Although  this 
reduces  slightly  the  time  spent  over  the  drawing  boards  the  work 

iii 


271051 


IV  PREFACE 

accomplished  by  the  students  is  not  apparently  diminished  and 
the  work  seems  to  be  generally  better  understood. 

Although  intended  primarily  for  a  textbook,  it  is  hoped  that 
this  may  prove  a  satisfactory  reference  book  for  designers  whose 
work  is  not  too  highly  specialized. 

The  author  desires  to  acknowledge  his  indebtedness  to  many 
manufacturers,  all  of  whom  have  been  most  generous  and  among 
whom  are  American  Bridge  Co.,  Pennsylvania  Steel  Co.,  Cam- 
bria Steel  Co.,  Bethlehem  Steel  Co.,  McClintock-Marshall  Con- 
struction Co.,  Jeffrey  Mfg.  Co.,  Best  &  Co.,  and  Bucyrus  Co. 

Among  the  periodicals  he  is  especially  indebted  to  the 
Engineering  News. 

The  section  on  Specifications  has  been  drawn  largely  from  the 
specifications  of  the  American  Railway  Engineering  and  Main- 
tenance of  Way  Association,  the  specifications  of  Mr.  C.  C. 
Schneider  in  the  Transactions  of  the  American  Society  of  Civil 
Engineers,  and  the  specifications  of  the  American  Bridge  Co. 

H.  D.  HESS. 

ITHACA,  N.  Y. 
May,  1913. 


CONTENTS 


CHAPTER  I. 
MATERIALS. 

Cast  Iron  —  Steel  —  Steel  Castings  —  Physical  Properties  of  Metals  —  Proper- 
ties of  Timbers  —  Timber  Beams  —  Timber  Columns  —  Bricks  —  Table  of 
Bending  Moments,  Deflections,  Etc.  —  Table  of  Properties  of  Sections  — 
Strength  of  Flat  Plates  —  Properties  of  Sections  of  Rolled  Structural  Steel.  1-19 

CHAPTER  II. 
GRAPHICS. 

Force  —  Magnitude,  Direction  and  Line  of  Action  of  a  Force  —  Equilibrium  — 
Couple  —  Resultant  —  Equilibrant  —  Force  Triangle  —  Force  Polygon  - 
Components  —  Composition  of  Forces  —  Resolution  of  Forces  —  Graphic 
Moments  —  Moments  of  Parallel  Forces  —  Uses  of  Force  and  Equilibrium 
Polygons  —  Graphical  Determination  of  Deflections  —  Graphical  Analyses  of 
Restrained  and  Continuous  Beams 20-34 

CHAPTER  III. 
STRESSES  IN   STRUCTURES. 

Tension  or  Compression  in  a  Member  —  Stresses  due  to  Moving  Loads  —  Wind- 
load  Stresses  —  Maximum  Bending  due  to  Moving  Loads  —  Maximum  Live- 
load  Shears 35~5 l 

CHAPTER  IV. 
ALGEBRAIC  DETERMINATION   OF  STRESSES. 

Method  of  Moments  —  Method  of  Coefficients 52-60 

CHAPTER  V. 
INFLUENCE   DIAGRAMS. 

Position  of  Loading  for  Maximum  Moments  —  Position  of  Loading  for  Maximum 
Shear  —  Position  of  Loading  for  Maximum  Floor-beam  Reaction  —  Moment 
Table  for  Cooper's  Loading,  E-6o 61-75 


vi  CONTENTS 

CHAPTER   VI. 
TENSION  PIECES,  COMPRESSION  PIECES  AND  BEAMS. 

Tension  Pieces  —  Compression  Pieces  —  Beams  —  Vertical  Shear  —  Bending 
Moment  —  Standard  Framing  —  Riveting  —  Shearing  and  Bearing  Value  of 
Rivets  —  Rules  for  Spacing  Rivets 76-89 

CHAPTER  VII. 
COLUMNS. 

Column  Formulae  —  Combined  Stresses  —  Long  Beams  Unsupported  Laterally. 

90-100 

CHAPTER  VIII. 
GIRDERS  FOR  CONVEYORS. 

Dead  and  Live  Loading  —  Wind  Loading  —  Determination  of  Stresses  —  Selection 
of  Members 101-109 

CHAPTER  IX. 
TRUSSES,   BENTS    AND     TOWERS. 

Trusses  and  Bents  for  carrying  Pipes  —  Towers  for  Electrical  Transmission 
Lines 1 10-1 1 7 

CHAPTER  X. 
DESIGN  OF  STEEL  MILL  BUILDINGS. 

Loading  —  Bracing  —  Determination  of  Stresses  —  Selection  of  Members  .118-140 

CHAPTER  XI. 
DESIGN   OF  A  PLATE-GIRDER  RAILWAY  BRIDGE. 

Loading  —  Bending  Moments  —  Vertical  Shears  —  Flange  Area  and  Selection 
of  Sections  —  Lengths  of  Flange  Plates  —  Spacing  of  Flange  Rivets  —  Wind 
Bracing  —  End  Diagonals  —  Web  Splice  —  Flange  Splice 141-161 

CHAPTER  XII. 
CRANE  FRAMES. 

Design  of  a  Frame  for  an  Underbraced  Jib  Crane  —  Design  of  a  Frame  for  a 
Top-braced  Jib  Crane 162-173 


CONTENTS  Vii 

•     f 

CHAPTER  XIII. 
GIRDERS   FOR   OVERHEAD   ELECTRIC   TRAVELING   CRANES. 

Box  Girders  —  Plate  Girders  with  Upper  Flange  Stiffened  by  a  Channel  —  Bridge 
Girders  with  Horizontal  Stiffening  Girders 174-190 

CHAPTER  XIV. 
REINFORCED   CONCRETE. 

Cement  —  Sand  —  Stone  and  Gravel  —  Physical  Properties  of  Concrete  —  Unit 
Working  Stresses  —  Rectangular  Beams  —  Approximate  Formulae  —  T  Beams  — 
Beams  with  Double  Reinforcing  —  Web  Stresses  —  Bond  Stresses  —  Lengths 
of  Reinforcing  Rods  —  Bent  Bars  —  Web  Reinforcements  —  Columns.  191-224 

CHAPTER  XV. 
FOUNDATIONS. 

Materials  —  Allowable  Pressure  on  Soils  —  Machine  Foundations  —  Building 
Foundations  —  Piles 225-237 

CHAPTER  XVI. 
CHIMNEYS. 

Kern  of  a  Section  —  Design  of  a  Brick  Chimney  —  Design  of  a  Self-sustaining 
Steel  Chimney  —  Design  of  a  Reinforced-concrete  Chimney 238-272 

CHAPTER  XVII. 
RETAINING  WALLS. 

General  Theory  of  Retaining  Walls  — Angles  of  Repose  of  Materials  —  Design 
of  Concrete  Retaining  Walls  —  Design  of  Reinforced-concrete  Retaining 
Walls 273-288 

CHAPTER  XVIII. 
BINS. 

Determination  of  Pressures  on  the  Sides  and  Bottoms  of  Bins  —  Stresses  in  Bins  — 
Graphical  Determination  of  the  Forces  acting  on  Bins  —  Hopper  Bins  — 
Suspension  Bunkers  —  Bin  Design 289-313 


viii  CONTENTS 

CHAPTER  XIX. 
SHOP  FLOORS. 

Cement  Concrete  —  Wooden  Blocks  —  Asphalt  —  Brick  —  Wooden  —  Upper 
Floors,  Wooden  —  Steel  —  Brick  Arch  —  Reinforced  Concrete  —  Iron.  314-322 

CHAPTER  XX. 
WALLS  AND   ROOFS. 

Wooden  Sides  —  Corrugated  Steel  Sides  — Walls,  Brick  —  Stone  —  Solid  Con- 
crete —  Hollow  Concrete  —  Reinforced  Concrete  —  Windows  —  Roof  Cover- 
ings, Corrugated  Steel  —  Slate  —  Clay  Tiles  —  Concrete  —  Slag  or  Gravel. 

323-343 

CHAPTER  XXI. 
SPECIFICATIONS. 

General,  Materials  —  Workmanship  —  Painting  —  Inspection  —  Testing  —  Spe- 
cifications for  Steel  Mill  Buildings  —  Specifications  for  a  Deck  Plate-girder 
Railway  Bridge  —  Specifications  for  Reinforced  Concrete 344-377 

CHAPTER  XXII. 
Problems 378-413 


GRAPHICS 
AND   STRUCTURAL   DESIGN 


CHAPTER   I 
MATERIALS 

THE  principal  materials  used  in  the  structures  considered  in 
this  book  are  iron  castings,  hard,  medium  and  soft  steels,  steel 
castings,  timber  (oak,  hemlock,  long-leaf  Southern  and  yellow 
pine,  spruce  and  white  pine)  and  concrete,  cement,  slag  and 
stone. 

Cast  Iron  is  made  by  remelting  pig  iron  or  pig  iron  and  cast- 
iron  scrap  in  a  cupola.  The  quality  of  the  product  is  a  varying 
one,  as  the  entire  charge  in  the  cupola  is  never  melted  at  any 
one  time.  The  quality,  therefore,  depends  upon  the  time  at 
which  it  was  run  off  during  the  heat. 

For  machining,  cast  iron  should  be  soft,  and  should  have  an 
ultimate  strength  in  tension  of  from  16,000  to  20,000  Ibs.  per 
sq.  in.  Although  cast  iron  has  no  well-defined  elastic  limit,  it 
may  be  assumed  for  practical  purposes  at  8000  Ibs.  per  sq.  in. 
Cast  iron  is  exceedingly  strong  in  direct  compression;  when  no 
bending  is  introduced  its  ultimate  strength  in  compression  should 
reach  90,000  to  100,000  Ibs.  per  sq.  in.  The  resilience  of  cast 
iron  being  very  low  indicates  little  ability  to  resist  shock. 

Steel  is  made  by  refining  pig  iron  in  the  Bessemer  converter, 
also  by  refining  pig  iron  or  pig  iron  and  steel  scrap  in  the  open- 
hearth  furnace.  Small  quantities  of  special  steels  are  also  being 
made  in  electric  furnaces.  The  refining  processes  remove  the  im- 
purities and  the  carbon,  the  steel  being  afterwards  recarbonized. 
The  ordinary  merchant  and  structural  steels  are  soft  and  medium 


i          O    r-     f 


2  GRAPHICS   AND   STRUCTURAL   DESIGN 

steels,  the  carbon  contents  ranging  from  0.08  to  0.30  of  i  per 
cent.  The  ultimate  strength  of  steel  varies  with  the  percentage 
of  carbon  from  48,000  Ibs.  per  sq.  in.  for  very  soft  or  rivet  steel 
to  70,000  Ibs.  per  sq.  in.  for  medium  steel.  Steel,  having  great 
resilience,  is  peculiarly  adapted  to  resisting  shock. 

Steel  Castings.  — •  This  steel  is  also  made  in  open-hearth 
furnaces  but  may  be  made  in  small  converters.  The  steel  is 
poured  into  molds  as  is  done  for  iron  castings.  To  obtain 
fluidity  it  is  necessary  that  steel  for  castings  have  a  much  higher 
temperature  than  is  required  for  cast  iron.  The  mold  frames 
must,  therefore,  be  much  more  substantial  and  the  molds  must 
be  thoroughly  dried  before  the  metal  is  poured  into  them. 

Considerable  contraction,  with  the  attendant  internal  stresses, 
results  from  the  exceedingly  high  temperature  and  necessitates 
the  annealing  of  steel  castings.  The  ultimate  strength  depends 
upon  the  carbon  and  should  be  50,000  or  more  pounds  per 
square  inch.  Steel  castings  have  high  resilience  and  are,  there- 
fore, superior  to  iron  castings.  However,  both  the  material 
and  the  machining  of  steel  castings  cost  more  than  do  those 
of  iron  castings. 

TABLE  OF  PHYSICAL  PROPERTIES  OF  METALS 


Material. 

Modulus  of  elasticity. 

Ultimate  strength. 

Elastic  strength. 

Tension 

compres- 
sion. 

Torsion. 

Tension. 

Shear. 

Tension. 

Shear. 

Cast  iron  (cupola)  

{10,700,000 
15,000,000 

28,000,000 

30,000,000 
30,000,000 
31,000,000 

30,600,000 

4,000,000 
6,000,000 

11,000,000 

11,800,000 
11,800,000 

12,100,000 

11,800,000 

16,000 

20,000 

30,000 
40,000 
{  47,000 
\  57,000 
60,000 
70,000 
100,000 
(    50,000 
(  100,000 

16,000  ) 

20,000  ) 

8,000 

8,000 

Wrought  iron  

35,000 
43,000 
45,000 
52,000 

30,000 
60,000 

40,000 
45.000 
75,ooo 

Steel  0.15  carbon  
Steel  0.25  carbon  
Crucible  steel  (high  carbon)  .  .  . 
Steel  castings 

NOTE.  —  The  ultimate  compressive  strength  of  cast  iron  is  90,000  to 
100,000  Ibs.  per  sq.  in.  Its  elastic  strength  in  compression  may  be  taken 
at  25,000  Ibs.  per  sq.  in.  The  ultimate  compressive  strengths  of  the  other 
materials  will  approximate  their  ultimate  tensile  strengths. 


MATERIALS  3 

Working  Fiber  Stresses.  —  In  the  structures  hereinafter  de- 
scribed the  usual  working  fiber  stresses  will  be  given  in  each 
problem.  Ordinarily,  in  structures  liable  to  little  or  no  shock 
or  vibration,  where  the  stresses  are  fully  determined,  the 
maximum  fiber  stress  on  mild  steel  will  range  from  16,000  to 
20,000  Ibs.  per  sq.  in.  Structures,  such  as  crane  frames,  liable 
to  some  shock  will  have  the  maximum  stress  on  mild  steel 
reduced  to  11,000  or  12,000  Ibs.  per  sq.  in.  In  the  case  of  high- 
way bridges  and  similar  structures  an  addition  of  25  per  cent  is 
added  to  the  live-load  stresses  to  allow  for  impact,  and  a  unit 
stress  of  15,000  Ibs.  per  sq.  in.  is  allowed  on  soft  steel,  while 
a  unit  stress  of  17,000  Ibs.  per  sq.  in.  is  allowed  on  medium 
steel.  In  the  case  of  railway  bridges  15,000  and  17,000  Ibs. 
per  sq.  in.  are  used,  but  the  impact  allowance  is  made  by  a 
formula  similar  to  that  used  in  the  design  of  the  railway  plate 
girder  in  Chapter  XI. 

PROPERTIES  OF  TIMBER 

Wooden  beams  are  designed  similarly  to  metal  ones.  Being 
liable  to  fail  by  horizontal  shearing  they  should  be  examined  for 
this.  One  -fifth  the  ultimate  shearing  resistance  given  in  the  tables 
may  be  taken  as  the  working  shearing  strength  (lengthwise). 

The  formula  for  bending  is 

„-/- 

M  =  bending  moment  in  inch  pounds. 

/    =  moment  of  inertia  in  inches. 

e    =  distance  from  the  neutral  axis  to  the  extreme  fibers  in 

inches. 

/    =  working  fiber  stress,  in  flexure,  pounds  per  square  inch. 
For  rectangular  timber  beams  this  becomes 


6 

b  =  width  of  the  beam  in  inches. 
d  =  depth  of  the  beam  in  inches. 


GRAPHICS  AND   STRUCTURAL  DESIGN 


The  maximum  unit  shearing  stress  is/,  =    -^  ,  where 

2  oa 

fa  =  fiber  stress  in  shear,  pounds  per  square  inch. 
R  =  end  reaction  in  pounds. 

TIMBER  COLUMNS 

The  following   formula  is  suggested  by  the  United  States 
Government  reports  on  timber. 

/  =F  x 


fc  =  ultimate  compressive  strength  of  the  column  in  pounds 

per  square  inch. 
Fc  =  ultimate  crushing  strength  of  short  timber  column  in 

pounds  per  square  inch. 

c  =  d' 

where 

/  =  length  of  the  column  in  inches. 
d  =  least  diameter  in  inches. 

The  factor  of  safety  should  vary  from  5  for  18  per  cent  moisture 
when  used  in  the  open  to  3^  for  10  per  cent  or  less  of  moisture 
when  used  in  heated  buildings. 


Name  of  wood. 

Pounds  per 
square  inch. 

White  oak,  Southern  long-leaf  pine  

<?OOO 

Short-leaf  yellow  pine  (Georgia) 

4.CQO 

Hemlock,  chestnut  and  spruce 

4.OOO 

White  pine  and  cedar  

3500 

The  properties  of  concrete  will  be  treated  quite  fully  under 
Reinforced  Concrete;  see  Chapter  XIV. 

Bricks.  —  Bricks  will  vary  in  size  and  properties  according 
to  the  locality  in  which  they  are  made.  Common  bricks  may 
generally  be  assumed  about  8j  ins.  X  4  ins.  X  2j  ins.,  while  face 
bricks  will  run  8f  ins.  X  4!  ins.  X  2\  ins.  Common  bricks  will 


MATERIALS 


tn 

& 
W 

- 
§ 

H 

O 
<n 

a 

0 

a 

H 

•^ooj  oiqno  J3d 
spunod  '^3pAV 

QS  r«0  M    <N    to  O    **•        OO      •    COO    O 

BASIS.  —  United  States  Department  of  Agriculture,  Forestry  Division,  moisture  12  per  cent  or  less. 
All  quantities  are  in  pounds  per  square  inch. 

Ordinary  working 
stresses  . 

g  s 

8888  &8  8     8:888 

M  00    O\  O   t^  to  O        to    -<N^OO\ 

B 

o. 
O 

§888     88     8:888 

ooot^      o  ^~      O    -o\r^t^ 

31 

§888     88     8  :888 

<vT    M"    M      M               MM*"             M"        :     M      M"    M 

Modulus  of 

Elastic 
bending. 

88:8     88     8^8:8 

ooq_    •  •<£     o_  •*      ^2^  *t    !    - 

Ultimate 
bending. 

8888888     88888 

OO    CO  M    O  M    M    O        O  O    CO  O    O 

OOOO    t^cor-         cs    COMOO    0 

|| 

OOOO     -QQ       QQQ2Q 
OOOO        QQ        QQQQQ 
OOOO        OO        OOOOO 

oooo      o  o      o"cTooo 

M               MM               CSM               <NCN«NMM 

II 

88  :8     88     888  :8 

OOO      •   •<*•       O   ^t        O    M    o«      •   Tt 

t^.  to   .  o       oo       o  M  o^   .  oo 

M     M 

o         w 

1 

O  ^ 

ojcoto-t^-=i-to      o     •     •  «     • 

VQMM        .M^-CN             tO.        .<O. 

II 

OO        OOQQ        ""'OOQO 
O   O         OOOO         co  f~**  O   O   O 

M                                           M 

Ultimate  resista 
Compression. 

II 

88     8     88    38S>:8 

Cht^          00              <Nt^            (N<NM-C^ 

Length- 
wise. 

8888888     §88§8 

t>.  to  to  to  tooo  to      oo  oo  1^0  t^ 

H 

§-OQQQO        QQQQQ 
§>§§.!§     §8888 

t^         M    coco"  co  O~          CO  CO  co  M    oT 

Name  of  wood. 

:      :  :  :  :  :  ^  :  :      :  : 

o    •    •       •    • 

'oj     '     '          '     ' 

.     .     .O    ' 

jifilPMff 

6  GRAPHICS  AND   STRUCTURAL  DESIGN 

weigh  125  Ibs.  per  cu.  ft.,  while  soft  bricks  will  average  100  Ibs. 
per  cu.  ft. 

Good  bricks  should  be  sound,  hard,  of  regular  shape  and  size, 
and  should  have  a  uniform  structure.  When  struck  a  sharp  blow 
they  should  emit  a  ringing  sound.  They  should  not  absorb  more 
than  TV  their  weight  of  moisture.  Their  crushing  strength 
should  reach  7000  to  10,000  Ibs.  per  sq.  in. 

Fire  Bricks.  —  These  are  bricks  made  of  clay  or  of  a  mixture 
of  clay  and  sand  having  ability  to  resist  high  temperatures. 
They  are  used  mainly  for  building  furnaces  and  lining  flues  and 
some  chimneys. 


MATERIALS 


BENDING  MOMENTS,  DEFLECTIONS,   ETC.,   FOR  BEAMS  OF 
UNIFORM  SECTION 


Form  of  support  and 
load. 


End  reactions  A  ,  B. 
Bending  moment  M  . 


Relation  between 
load  "  P  "  and  mo- 
ment of  resistance. 


Maximum  deflection. 


D  =:  i    • 

A7=Px, 


'    / 
•1F-5?, 


2 
P/ 

4 


d=—.- 


P     /' 


Mmax 


L    '        "T 
Pec1 
/ 


ccl 


4=£  =  P, 

Between  A  &  B, 
M=P-c  =  const. 


_P_/3c 
£78]' 

8£V 

P  /c3   .   c2/ 


M  Pl 

-Mmax  =  — 


£7*8' 


p 


48  Ee 


GRAPHICS   AND    STRUCTURAL   DESIGN 


BENDING  MOMENTS,   DEFLECTIONS,   ETC.,   FOR  BEAMS   OF 
UNIFORM  SECTION 


Form  of  support. 


End  reactions  A  ,  B. 
Bending  moment. 


Relation  between 
load  "  P  "  and  mo- 
ment of  resistance. 


Maximum  deflection. 


=  -P,   B  =  -  P, 


fW 

P  =    -*— 


w  = 


Pl 

7-794/' 


PI3 
d  =  0.01304-^ 


A  =  B  =  -, 


Pl 

— 
O 


60  £/ 


_L/L2 
10  Ee 


M=  — 


r3_*\ 
,4        l) 


r 


8fW 

~> 


--- 


185  £/ 


. 

id 


, 

16  16 

M  3P/ 

ikZmax  —  —  7—' 
16 


16  fW 
"  3       ^ 

IT-A«. 

16    / 


PI3 


A  =  B  =  P, 


A/max  ==  —  * 


p  =  ^jwi 


d  = 


PI* 


i    //* 
32  Ef 


5  =  -, 

2 

=  ~T-  ' 
o 


P  = 


7-<   T- 

192  £/ 
24  Ee' 


V  *  '  / 


MATERIALS  g 

The  bending  moment  is  a  maximum  under  load  i  when  x  =  -• 

4 


Pi 


max 


The  letters  used  have  the  following  significance  and  for  con- 
venience they  should  be  expressed  in  the  units  stated. 
A  and  B  =  end  reactions,  e  =  distance  neutral  axis  to 

pounds.  extreme  fibers  having 

P  =  load,  pounds.  fiber  stress  /,  inches. 

/  =  extreme  fiber  stress,  E  =  modulus    of    elasticity, 

pounds  per  square  inch.  pounds  per  square  inch. 

/  =  span  of  beam  in  inches.        c,  Ci  and  x  =  portions  of  span, 

inches. 

W  =  resistance,  =  -  •  x  =  left  end  to  section  where 

bending  is  wanted,  inches. 
.  7  =  moment  of  inertia,  inches,     d  =  deflection,  inches. 

STRENGTH  OF  FLAT  PLATES 

Nomenclature : 

t  =  thickness  of  plate  in  inches. 

/„  =  maximum  fiber  stress  in  plate,  pounds  per  square  inch. 
E  =  modulus  of  elasticity  in  flexure,  pounds  per  square  inch. 
c  factor  according  to  Grashof  or  Bach. 
P  =  concentrated  load  in  pounds,  usually  assumed  as  acting 

on  a  circle  whose  radius  is  r0  ins. 

q  =  uniform  load  acting  on  plate,  pounds  per  square  inch. 
r  =.  radius  of  circular  plate,  inches. 
b  =  longer  side  of  rectangular  plate,  inches. 
a  =  shorter  side  of  rectangular  plate,  inches. 
b  _ 

~  —    €. 

a 

i.   Circular  plates,  carrying  uniform  load  q. 

r2 


IO 


GRAPHICS  AND   STRUCTURAL  DESIGN 


MOMENTS  OF  INERTIA,  RESISTANCES,  CENTERS  OF  GRAVITY 

AND  LEAST  RADII  OF  GYRATION  OF 

GEOMETRICAL  SECTIONS 


Shape  of 
section. 


Moment  of  inertia. 


Resistance. 


Distance  base 

to  center  of 

gravity. 


Least  radius  of 
gyration. 


W 
6 


Lesser  side 
3-46 


6      B 


! 


bh* 
36' 


bh* 

24' 


The  smaller 

A  b 

-    or    -- 
4.24          4.9 


36(26+61) 


3      2&  +  &J 


20.38 


D3 
10.19' 


0.049  (D4— ( 


0.098 


Wz=  0.259  R3. 


0.424  R. 


0.07  £2 


0.7854^2. 


MATERIALS  II 

Cast-iron  plate  supported  at  outer  circumference,  c  =  0.8  to  1.2. 
Steel  plate  supported  at  outer  circumference,          c  =  0.75. 
Steel  plate  bolted  or  riveted  at  outer  circumference,  c  =  0.50. 

2.   Circular  plates,  carrying  a  central  load  P,  applied  on  a  circle 
of  radius  r0.    Plate  supported  at  outer  circumference. 


For  cast  iron  c  =  1.5. 

3.   Circular  plates,  central  load  similar  to  2  but  plate  bolted 
or  riveted  at  outer  circumference. 


RECTANGULAR  AND  SQUARE  PLATES 

Rectangular  and  square  plates  supported  at  the  outer  edges 
and  carrying  a  uniform  load  q. 
Rectangular, 

V2 


Square, 

ca* 


For  cast  iron  c  =  0.75  to  1.13. 

For  steel         c  —  0.56  to  0.75  (max.  1.13). 


STRUCTURAL  MATERIAL 

The  principal  rolled  sections  used  are  I  beams,  channels, 
angles,  plates,  flats  and  rounds.  Manufacturers'  handbooks 
afford  the  best  sources  of  information  regarding  these  sections 
and  in  actual  design  they  should  be  freely  consulted.  The 
following  data  are  abridged  from  these  books. 


12 


GRAPHICS  AND   STRUCTURAL  DESIGN 


SHEARED  PLATES 


Width  in  inches. 

Thickness  in  inches. 

A 

i 

A 

1 

A 

i 

T9* 

t 

H 

1 

it 

1 

if 

Length  in  inches. 

IS 

240 

320 

400 

500 

500 

550 

500 

475 

475 

475 

425 

400 

375 

360 

300 

280 

280 

16 

240 

320 

400 

500 

500 

550 

500 

475 

475 

475 

425 

400 

375 

360 

300 

280 

280 

17 

240 

320 

400 

500 

500 

550 

500 

475 

475 

475 

425 

400 

375 

360 

300 

280 

280 

18 

240 

360 

400 

500 

500 

500 

500 

550 

550 

550 

500 

500 

450 

400 

400 

400 

350 

iQ 

240 

360 

400 

500 

500 

500 

500 

55° 

550 

55° 

500 

500 

450 

400 

400 

400 

350 

20 

240 

360 

400 

500 

500 

500 

500 

550 

550 

550 

500 

500 

450 

400 

400 

400 

350 

21 

216 

360 

400 

500 

525 

525 

525 

55° 

550 

550 

500 

500 

450 

400 

400 

400 

350 

22 

216 

36o 

400 

500 

525 

525 

525 

550 

550 

550 

500 

500 

45° 

400 

400 

400 

350 

23 

204 

36o 

400 

500 

525 

525 

525 

550 

550 

550 

500 

500 

450 

400 

400 

400 

350 

24 

204 

360 

400 

500 

525 

550 

55° 

550 

550 

550 

500 

475 

425 

400 

35° 

350 

325 

25* 

204 

360 

400 

500 

525 

550 

550 

55° 

550 

550 

500 

475 

425 

400 

350 

350 

325 

26 

1  80 

360 

400 

500 

525 

550 

550 

55° 

550 

55° 

500 

475 

425 

400 

350 

350 

325 

2? 

168 

340 

400 

500 

500 

550 

550 

500 

500 

500 

45° 

450 

400 

380 

330 

300 

300 

28 

168 

340 

400 

500 

500 

550 

550 

500 

500 

500 

450 

45° 

400 

380 

330 

300 

300 

29 

156 

340 

400 

500 

500 

550 

550 

500 

500 

500 

450 

450 

400 

380 

330 

300 

300 

30-  35 

320 

400 

500 

500 

550 

500 

475 

475 

475 

425 

400 

375 

360 

300 

280 

280 

36-  41 

360 

400 

500 

500 

500 

500 

550 

55° 

550 

500 

500 

450 

400 

400 

400 

350 

42-  47 

360 

400 

500 

525 

525 

525 

55° 

55° 

550 

500 

.500 

450 

400 

400 

400 

350 

48-  53 

360 

400 

500 

525 

550 

55° 

55° 

550 

550 

500 

475 

425 

400 

350 

350 

325 

54-  59 

340 

400 

500 

500 

550 

550 

500 

500 

500 

450 

450 

400 

380 

330 

300 

300 

60-  65 

320 

400 

500 

500 

550 

500 

475 

475 

475 

425 

400 

375 

360 

300 

280 

280 

66-  71 

300 

350 

430 

450 

475 

425 

425 

425 

410 

375 

340 

330 

320 

280 

260 

260 

72-  77 

260 

300 

400 

425 

450 

400 

400 

400 

390 

350 

320 

320 

300 

260 

240 

240 

78-83 

240 

275 

380 

400 

420 

375 

375 

375 

37° 

325 

300 

300 

300 

240 

220 

2  2O 

84-  89 

200 

250 

35° 

375 

385 

35° 

350 

350 

350 

300 

280 

2/5 

275 

230 

2IO 

2IO 

90-  95 

1  80 

230 

330 

340 

350 

350 

325 

325 

325 

275 

260 

260 

260 

220 

200 

2OO 

96-101 

120 

175 

240 

250 

275 

275 

275 

275 

275 

240 

240 

220 

220 

2OO 

1  80 

1  80 

102-107 

150 

200 

230 

230 

250 

250 

250 

250 

230 

230 

2IO 

210 

190 

170 

170 

108—113 

1  80 

1  80 

200 

220 

22=; 

22$ 

22< 

220 

220 

2OO 

2OO 

1  80 

160 

160 

114—119 

1  80 

200 

•*O 

2IO 

*•  ^0 
2IO 

•*o 

2IO 

200 

200 

1  80 

1  80 

I7O 

I  ^O 

I  SO 

120-125 

I2O 

150 

150 

1  80 

1  80 

i75 

175 

160 

1  60 

160 

*  ov 

144 

*  ow 
144 

MATERIALS 
EDGED  PLATES 


Thickness  in  inches. 

Width  in 
inches. 

A 

1 

4 

A 

1 

T% 

» 

A 

| 

H 

3. 

4 

11 

I 

« 

I 

Length  in  feet. 

4 

50 

50 

50 

50 

50 

5° 

40 

40 

30 

30 

3° 

28 

28 

28 

5 

30 

42 

42 

42 

42 

40 

30 

30 

30 

30 

30 

30 

30 

30 

6 

3° 

42 

42 

42 

42 

40 

35 

30 

30 

3° 

30 

30 

30 

30 

7 

25 

42 

42 

42 

42 

40 

35 

30 

30 

3° 

30 

30 

30 

30 

8 

25 

42 

42 

42 

42 

42 

38 

36 

32 

3° 

29 

28 

26 

25 

9 

25 

42 

42 

42 

42 

42 

38 

34 

32 

30 

29 

28 

26 

25 

10 

25 

42 

42 

42 

42 

42 

38 

33 

32 

3° 

29 

28 

26 

25 

ii 

25 

42 

42 

42 

42 

42 

38 

33 

31 

29 

28 

27 

25 

24 

12 

25 

42 

42 

42 

42 

42 

37 

32 

30 

28 

27 

26 

24 

23 

13 

42 

42 

42 

42 

42 

37 

32 

30 

27 

25 

24 

22 

2O 

14 

42 

42 

42 

42 

40 

35 

30 

28 

26 

25 

23 

22 

2O 

14* 

42 

42 

42 

42 

36 

33 

30 

28 

25 

In  the  tables  of  angles  the  areas  are  given,  and  the  weights  in 
pounds  per  foot  of  section  can  be  obtained  by  multiplying  the 
areas  by  3.4. 

The  moments  of  inertia  and  location  of  the  centers  of  gravity 
of  the  angles  are  given  in  the  tables  following. 

The  radii  of  gyration  are  readily  found  from 


I  =  inertia; 
A  = 


area  in  square  inches. 

All  data  are  based  upon  dimensions  in  inches 
pounds. 


and  weights  in 


GRAPHICS  AND   STRUCTURAL  DESIGN 


DIMENSIONS  AND   AREAS  OF   ANGLES 


Uneven 
legs. 

Even 
legs. 

1 

1% 

1 

A 

I 

T\ 

i 

A 

I 

H 

I 

H 

1 

it 

I 

4 



8  X8 
6  X6 

*5  X5 

5'o6 
4-40 
4.18 
3-97 
3-75 
3-53 
3-31 
3-09 
2.87 
2.65 
2-43 

2.22 
2.OO 

1.78 
1.56 
1.30 

7.75 
5.75 
5.00 
4.75 
4.50 
4.25 

4.00 
3-75 
3-50 
3-25 
3.00 
2.75 
2.50 
2.25 

2.00 

8.68 
6.43 
5-59 
5-31 
5.03 
4-75 
4-47 
4.18 
3-90 
3.62 
3-34 
3.06 
2.78 

9.61 
7-  II 
6.17 
5.86 
5-55 
5-23 
4.92 
4.61 
4.30 
3.98 
3.67 
3.36 

10.5 
7-78 
6.75 
6.41 
6.06 
5-72 
5-37 
5-03 
4.68 
4-34 
4.00 
3-65 

II.  4 
8.44 
7-31 
6.94 
6.56 
6.19 
5.8i 
5-44 
5.06 
4-69 
4-31 

12.3 

9.09 
7.87 

7-47 
7.06 
6.65 
6.25 
5-84 
5-43 
5.03 
4.62 

13.2 

9-74 
8.42 
7-99 
7-55 
7.11 
6.67 

14.1 
10.4 
8.97 
8.50 
8.03 

IS.O 
II.  0 
9-50 
9.OO 
8.50 

4.36 

3'6i 
3-42 
3-23 
3-05 
2.86 
2.67 
2.48 
2.30 

2.  II 
.92 

•  73 
•  55 
.36 
•  17 
0.99 

*7  X3i 
6  X4 
6  X3* 
5  X4 
5  X3* 
5  X3 
*4  X3* 
4  X3 
3*X3 
3*X2* 
3  X2* 
3  X2 
2^X2 

'56 
.40 
25 

4  X4 
'3*X3i 
'3'X3' 

*2fX2| 

2*X2* 

*2jX2j 
2    X2 

ijXil 
i*Xii 

1.44 
1.31 
1.19 
1.  06 
0.94 
0.81 
0.69 

.09 

•  93 
-78 
.62 
1-47 
I.3I 
1.  15 

I.OO 

0.84 

o'.36 

0.81 
0.72 
0.62 
0.53 

Special  sections. 


MOMENTS    OF     INERTIA    AND    CENTERS     OF     GRAVITY    OF 

ANGLES 

EVEN  LEGS 


Thick- 
ness. 

1 

1 

I 

{ 

I 

* 

I 

Legs. 

I 

g 

/ 

g 

I 

g 

I 

g 

/ 

g 

7 

g 

/ 

g 

8  X8 
6  X6 

15  4 

i  6 

48.6 

19  9 

.2 
7 

59-4 

24  2 

2.2 

i  7 

69.7 

28  2 

2.3 
I  8 

79-6 
31  9 

2.3 

r  8 

89.0 
35-5 

2.4 
1.9 

*5  X5 

8  7 

I  4 

II  3 

13  6 

I  5 

4  X4 
3*X3* 
3  X3 

I   2 

o  84 

4-4 

2-9 

I  8 

i.i 

I.O 

o  89 

5.6 
3.6 

2    2 

.2 
.1 

o  93 

6.7 
4.3 
2  6 

1.2 

I.I 

o  98 

7-7 

S.o 

1-3 

1.2 

8.6 
5-5 

1.3 
1.2 



"2fX2f 

O.95 

0.78 

1.3 

0.82 

I    7 

o  87 

2kX2k 

o  70 

o  72 

o  98 

o  76 

I    2 

o  81 

2iX2j 

o  50 

o  65 

o  70 

o  70 

2   X2 

O.35 

0.59 

0.48 

0.64 

O.59 

o  68 

Special  sections. 


MATERIALS 


MOMENTS    OF    INERTIA    AND    CENTERS    OF     GRAVITY    OF 

ANGLES 

UNEVEN  LEGS 


Thick- 
ness. 

Axis. 

i 

1 

i 

1 

* 

1 

I 

Legs. 

/ 

e 

/ 

g 

*7  X3i 
6  X4 
6  X3* 
*S  X4 
5  X3i 
5  X3 
*4  X3i 
4  X3 
3*X3 
3iX2j 
3  X2* 
3   X2 

2JX2 

i 

4-9 
13-5 
3-3 
12.9 

4-7 
8.1 

3-2 

7-8 

2.0 
7-4 

3-0 
4.2 
1.9 
4-0 
1.9 

2.7 
i.i 

2.6 
I.O 

1-7 
0.54 
i.S 
0.51 
0.91 

0.94 
1.94 
0.79 
2.0 
I.O 
1.5 
0.86 
1.6 
0.70 
1-7 
0.96 

1.2 

0.78 

1-3 
0.83 
i.i 
0.66 

1.2 
0.71 
0.96 

0.54 

I.O 

0.58 
0.83 

4.4 

lii 

17.4 
4.3 

16.6 
6.0 
10.5 
4-i 

10.  0 
2.6 

9-5 
3-8 
53 
2.4 
5.1 

2.3 

3-5 
1.4 

3-2 

1.3 

2.1 
0.67 
1-9 
0.64 
I.I 

0.78 
2.5 
0.09 
1.99 
0.83 
2.1 
I.I 

1.6 
0.91 
I  7 

?:25 

1.  00 
1.2 
0.83 
13 

0.88 
i.i 
0.70 

1.2 

0.75 

I.O 

0.58 

I.I 
0.63 
0.88 

5-3 
30.9 
7-5 

21.  1 

5-1 
20.1 
7-1 

12.6 

4-8 

12.0 

3-1 
II.  4 

ti 

1:1 

2.8 

4-1 
1.6 
3-9 
1-5 

2.5 

0.82 
.6 

.0 

:!i 

.1 
.1 
.6 
•  95 
.7 
.80 
.8 
.00 

'87 
•  4 
•  92 
.2 
•  75 
•  3 
•  79 

.0 

6.1 
36.0 
8.7 
24.5 
5.8 
23-3 

0.87 

2.6 

i.i 
2.1 
0.93 

2.2 

6.8 

40.8 
9-8 
27.7 
6.6 
26.4 

0.91 
2.7 
i.i 

2.1 
0.97 

2.2 

7-5 
45-4 
10.7 
30.8 

7-2 

29.2 

0.06 
2.7 
1.2 
2.2 
I.O 
2.3 

56 

I    O 

6  ? 

I   O 

13-9 
3-5 
13-2 

1.7 

0.84 
1.8 

15-7 
3-9 
14.8 

1.8 
0.88 
1.9 

'sii 

3.2 
4-9 
1.8 
4-4 

'78 
.8 
•  74 
.2 
•  39 
.1 
0.37 
0.65 

0.61 
i.i 

0.66 
0.91 
0-49 
0.99 
0.54 
0.79 

0.92 
1-4 
0.06 
1.2 
0.79 
1.3 

?:2 

3-5 
5-2 

0.96 
1.5 

I.O 

1-3 

Special  sections. 


The  properties  of  angles  of  intermediate  thickness  can  be 
interpolated  from  the  tables  with  sufficient  accuracy  for  most 
purposes. 


i6 


GRAPHICS  AND   STRUCTURAL  DESIGN 


PROPERTIES  OF   STANDARD  CHANNELS 


" 

Weight 
per 
foot. 

Flange, 
b. 

Web, 

Gauge, 
m. 

Tan- 
gent, 

Max. 

bolt  or 
rivet. 

Moment 
of  in- 
ertia, 
axis  i-  1. 

Moment 
of  in- 
ertia, 
axis  2-2. 

Dis- 
tance 
base 
toe. 

ofg. 

Area. 

Radius  of 
gyration. 

Axis 
i-i. 

Axis 

2-2. 

55-00 

3-82 

0.82 

2.50 

12.25 

1      r 

430 

12.2 

0.82 

16.18 

5-i6 

0.8? 

50.00 
45-00 

3-72 
3-62 

0.72 
0.62 

2.50 

2.00 

12.25 
12.25 

L 

403 
375 

II.  2 

10.3 

0.80 
0.79 

14.71 
13-24 

5.23 
5-32 

0.87 

0.88 

15' 

40.00 

3-52 

0.52 

2.OO 

12.25 

•  *  1 

348 

9-4 

0.78 

11.76 

5-44 

0.89 

35-00 

3-43 

0.43 

2.00 

12.25 

320 

8.5 

0-79 

10.29 

5-57 

0.91 

. 

33-00 

3-40 

0.40 

2.OO 

12.25 

J      I 

312 

8.2 

0-79 

9.90 

5-62 

0.91 

r 

40.00 
35-00 

3-42 
3-30 

0.76 
0.64 

2.OO 
2.00 

IO.OO 
10.00 

I      J 

197 
179 

6.6 
5-9 

0.72 
0.69 

11.76 
10.29 

4-09 
4.17 

0.75 
0.76 

I2< 

30.00 

3-17 

0.51 

1-75 

IO.OO 

[  i  4 

162 

5-2 

0.68 

8.82 

4-28 

0-77 

j 

25.00 

3-05 

0.39 

1-75 

IO.OO 

\      \ 

144 

4-5 

0.68 

'7.35 

4-43 

0.78 

I 

20.50 

2.94 

0.28 

1-75 

10.00 

J      I 

128 

3-9 

0.70 

6.03 

4.61 

0.81 

r 

35-00 

3-18 

0.82 

1-75 

8.25 

}r 

116 

4-7 

0.69 

10.29 

3-35 

0.67 

30.00 

3-04 

0.68 

1-75 

8.25 

103 

4-0 

0.65 

8.82 

3-42 

0.67 

icn 

25.00 

2.89 

0.63 

1-75 

8.25 

i  •{ 

91 

3-4 

0.62 

7-35 

3-52 

0.68 

1 

20.00 

2.74 

0.38 

1-50 

8.25 

} 

79 

2-9 

0.61 

5  '88 

3.66 

0.70 

L 

15.00 

2.60 

0.24 

1.50 

8.25 

[ 

6? 

2-3 

0.64 

4.46 

3.87 

0.72 

r 

25.00 

2.81 

0.61 

1.50 

7-25 

i      r 

71 

3-0 

0.62 

7-35 

3.10 

0.64 

9 

20.00 
I5.OO 

2.65 
2.49 

0.45 
0.29 

1-50 
1-38 

7.25 
7-25 

H 

61 
51 

2.5 

2.O 

0.58 
0-59 

5-88 
4-41 

3-21 
3-40 

0.65 
0.66 

L 

13-25 

2.43 

0.23 

1-38 

7-25 

J     I 

47 

1.8 

0.61 

3.89 

3-49 

0.67 

r 

21.25 

2.62 

0.58 

1-50 

6.25 

}r 

48 

2-3 

0-59 

6.25 

2.76 

0.60 

18.75 

2.53 

0.49 

i-5o 

6.25 

44 

2.O 

0-57 

5-51 

2.82 

0.60 

8i 

16.25 

2.44 

0.40 

1-50 

6.25 

i  •{ 

40 

1.8 

0.56 

4-78 

2.89 

0.61 

1 

13-75 

2.35 

0.31 

1-38 

6.25 

36 

1.6 

0.56 

4-04 

2.98 

0.62 

L 

11.25 

2.26 

0.22 

1.38 

6.25 

I 

32 

1.3 

0.58 

3-35 

3-10 

0.63 

r 

19-75 

2.51 

0.63 

1-50 

5-50 

i      r 

33 

1.8 

0.58 

5-8l 

-39 

0.56 

17-25 

2.41 

0.53 

1-50 

5-50 

30 

1.6 

0.55 

5-07 

-44 

0.56 

H 

14-75 

2.30 

0.42 

1-50 

5-50 

r  *  i 

27 

1-4 

0.53 

4-34 

-50 

0.57 

12.25 

2.  2O 

0.32 

1-25 

5-50 

24 

1.2 

0.53 

3.6o 

•  59 

0.57 

I 

9-75 

2.09 

0.21 

1.25 

5-50 

J      I 

21 

0.98 

0.55 

2.85 

.72 

0.59 

r 

15.50 

2.28 

0.56 

1-25 

4-50 

1      r 

19-5 

1.3 

0-55 

4-56 

-07 

0-53 

A  J 

13.00 

2.16 

0.44 

1.25 

4-50 

I     j 

17-3 

i.i 

0.52 

3-82 

.13 

0.53 

1 

10.50 

2.04 

0.32 

1-25 

4-50 

i  f  1 

IS-  1 

0.88 

0.50 

3-09 

.21 

0.53 

L 

8.00 

1.92 

O.2O 

1.  13 

4-50 

J      I 

13-0 

0.70 

0.52 

2.38 

•  34 

0.54 

11.50 

2.04 

0.48 

1.13 

3-75 

>     c 

10.4 

0.82 

0.51 

3-38 

-75 

o.49 

9.00 

1.89 

0.33 

1.  13 

3-75 

L  §  J 

8.9 

0.64 

0.48 

.65 

-83 

0.49 

6.50 

1-75 

0.19 

1.  13 

3-75 

J       I 

7-4 

0.48 

0.49 

-95 

•  95 

0.50 

f 

7.25 

1-73 

0.33 

I.OO 

2.75 

}r 

4.6 

0.44 

0.46 

-13 

-46 

0.46 

4i 

6.25 

1.65 

0.25 

I.OO 

•  75 

i    J 

4.2 

0.38 

0.46 

.84 

•  Si 

0.45 

I 

5-25 

1.58 

0.18 

I.OO 

•  75 

I 

3.8 

0.32 

0.46 

.55 

•  56 

0.45 

*{ 

6.00 
S.oo 
4.00 

i.  60 
1.50 
i.  41 

0.36 
0.26 
0.17 

0.88 
0.88 
0.88 

•  75 
•  75 
•75 

H 

2.1 

1.8 
1.6 

0.31 
0.25 

0.2O 

0.46 
0.44 
0.44 

.76 
•  47 
•  19 

.08 

.12 

•17 

0.42 
0.41 
0.41 

Note.  —  This  table  is  taken  from  the  handbook  of  the  Cambria  Steel  Co. 


MATERIALS 


PROPERTIES  OF  STANDARD  I  BEAMS 
"f 


Depth  of  i 
beam. 

Weight  per 
foot. 

Area  of 
section  . 

Thickness 
of  web. 

Width  of 
flange. 

Moment  of 
inertia, 
axis  i-i. 

Section 
modulus, 
axis  i-i. 

Radius  of 
gyration, 
axis  i-i. 

Moment  of 
inertia, 
axis  2-2. 

Radius  of 
Ky  ration, 
axis  2-2. 

i! 

d 

A 

t 

b 

7 

S 

r 

r 

r' 

n 

T 

Ins. 

Lbs. 

Sq.  ins. 

In. 

Ins. 

Ins.* 

Ins.' 

Ins. 

Ins.« 

In. 

Ins. 

Ins. 

Ins. 

3 

5-50 

1.63 

0.17 

2.33 

2.5 

1.7 

1.23 

0.46 

o.53l 

3 

6.50 

•  91 

0.26 

2.42 

2.7 

1.8 

1.  19 

0.53 

0-52^ 

i 

1/8 

iii 

3 

7-50 

.21 

0.36 

2.52 

2.9 

1-9 

i.  IS 

0.60 

0.52J 

4 

7-50 

.21 

0.19 

8.66 

6.0 

3-0 

1.64 

0.77 

0-591 

4 
4 

8.50 
9-50 

-50 

•  79 

0.26 
0.34 

2.73 
2.8l 

6.4 
6.7 

3-2 

34 

1.59 
1.54 

0.85 
0.93 

o.5Sl 
o.58f 

i 

ij 

211 

4 

10.50 

3-09 

0.41 

2.88 

7-1 

3-6 

1.52 

1.  01 

0.57J 

S 

9-75 

2.87 

0.21 

3  oo 

12.  1 

4-8 

2.05 

1.23 

o.6s1 

5 

12.25 

3.6o 

0.36 

3  IS 

13-6 

54 

1.94 

1.45 

0.63  !> 

} 

1} 

3i 

S 

14-75 

4-34 

0.50 

3.29 

IS-I 

6.1 

1.87 

1.70 

o.63j 

000 

12.25 
14-75 
17-25 

3.6i 
4-34 
5-07 

0.23 
0.35 

0.47 

3  33 
3-45 
3-57 

21.8 
24.0 
26.2 

1:1 

8.7 

2.46 
2.35 

2.27 

1.85 
2.09 
2.36 

0.72^ 
0.69^ 
0.68J 

f 

2 

4A 

7 
7 
7 

I5.OO 
17.50 
2O.OO 

4-42 

13 

0.25 
0-35 
0.46 

3.66 
3.76 
3-87 

36.2 
39-2 
42.2 

10.4 

II.  2 
12.  1 

2.86 
2.76 
2.68 

2.67 
2.94 
3-24 

0.781 
0.76^ 
0-74J 

f 

4 

5} 

8 

18.00 

S-33 

0.27 

4.00 

56.9 

14.2 

3-27 

3-78 

0.841 

8 
8 

20.25 

22.75 

5-96 
6.69 

0.35 
0.44 

4.08 
4-17 

60.2 
64.1 

16.0 

3.18 
3.10 

4.04 
4.36 

0.82! 
0.81  f 

i 

21 

6A 

8 

25.25 

7-43 

0.53 

4.26 

68.0 

17.0 

3.03 

4-71 

o.8oj 

9 

21.00 

6.31 

0.29 

4-33 

84-9 

18.9 

3.6? 

5.16 

o.ool 

9 

25.0O 

0.41 

4-45 

91.9 

20.4 

3-54 

5-65 

0.881 

i 

, 
* 

, 

9 

30.00 

8.82 

0.57 

4.61 

101  .9 

22.6 

3-40 

6.42 

0.85  f 

t 

7i« 

9 

35.OO 

10.29 

0.73 

4-77 

in.  8 

24.8 

3-30 

7-31 

0.84J 

10 

25.OO 

7-37 

0.31 

4.66 

122.  1 

24.4 

4.07 

6.89 

10 

30.OO 

8.82 

0.45 

4.80 

134-2 

26.8 

3-00 

7-65 

. 

« 

* 

-is 

10 

35-00 

10.29 

o.oo 

4-95 

146.4 

29-3 

3-77 

8.52 

* 

716 

10 

40.00 

11.76 

o.75 

S.io 

158.7 

31-7 

3-67 

9-50 

12 

31.50 

9.26 

0.35 

5-00 

215.8 

36.0 

4-83 

9-50 

i.  oil 

12 

35-OO 

10.29 

0.44 

5-09 

228.3 

38-0 

4-71 

10.07 

0-99^ 

I 

2\ 

9! 

12 

40.00 

11.76 

0.56 

5.21 

245-9 

41-0 

4-57 

10.95 

0.96J 

IS 

42.00 

12.48 

0.41 

5-50 

441-8 

58-9 

5-95 

14.62 

.081 

IS 

45.00 

13.24 

0.46 

5-55 

455-8 

60.8 

5-87 

15.09 

-07| 

IS 

50.00 

14-71 

0.56 

5.6S 

483.4 

64.5 

5  73 

16.04 

.04!- 

I 

3 

12* 

15 

55-00 

16.18 

0.66 

5-75 

Sii.o 

68.1 

5-62 

17.06 

.03! 

IS 

60.00 

17.65 

0.75 

5.84 

538.6 

71.8 

5-52 

18.17 

.OlJ 

18 

55-0 

15-93 

0.46 

6.00 

795-6 

88.4 

7-07 

21.19 

.151 

0.875 

18 

60.0 

17.65 

0.56 

6.10 

841.8 

93-5 

6.91 

22.38 

.13  I 

i 

18 

65.0 

19.12 

0.64 

6.18 

881.5 

97-9 

6.79 

23.47 

'"I 

i 

3i 

I5l38 

18 

70.0 

20.59 

0.72 

6.26 

921.2 

102.4 

6.69 

24.62 

•  09J 

i 

20 

65.0 

19.08 

0.50 

6.25 

1169.5 

117.0 

7.83 

27.86 

.2l1 

2O 

70.0 

20.59 

0.58 

6.33 

1219.8 

122.  0 

7-70 

29.04 

•  *9  r 

i 

3* 

l6J| 

20 

75-0 

22.06 

0.65 

6.40 

1268.8 

126.9 

7-58 

30.25 

.IJJ 

24 

80.0 

23.32 

0.50 

7.00 

2087.2 

173-9 

9.46 

42.86 

.361 

24 

85.0 

25.00 

0.57 

7.07 

2167.8 

180.7 

9-31 

44-35 

.33 

24 

90.0 

26.47 

0.63 

7-13 

2238.4 

186.5 

9.20 

45-70 

.31  y 

i 

4 

2oli 

24 

95-0 

27.94 

0.69 

7.19 

2309.0 

192.4 

9.09 

47-10 

.30 

24 

IOO.O 

29.41 

0.75 

7-25 

2379-6 

198.3 

8.99 

48.55 

.28J 

Note.  — This  table  is  taken  from  the  handbook  of  the  Cambria  Steel  Co. 


i8 


GRAPHICS  AND   STRUCTURAL  DESIGN 


GREY  MILL  SECTIONS 

The  Bethlehem  Steel  Company  on  their  Grey  Mill  are  able 
to  roll  sections  with  much  wider  flanges  than  are  possible  with 
the  usual  methods  of  manufacture.  These  sections  make  better 
columns  and  can  be  used  upon  longer  spans  without  lateral  brac- 
ing. The  following  tables  give  some  of  these  sections  with  their 
properties.  In  H  columns,  under  each  section-number,  only  the 
first  three  weights  and  the  maximum  weight  are  given  although  a 
large  number  of  intermediate  weights  are  rolled.  The  beam  and 
girder-beam  sections  are  given  almost  completely. 

H  COLUMNS 


| 

i 

Axis  XX. 

Axis  FF. 

£ 

"5 

I 

•M 

1 

P. 

C/3 

0 

D 

b 

t 

r 

"o 

gj 

"ogri 

15  jj 

§3 

Ofl     . 

MOW 

'S 

*r>  rj 

.2? 
1 

1 

a 

i| 

IP 

£ 

J.S 

l| 

IP 

Lbs. 

Sq.  ins. 

/ 

5 

r 

r 

S' 

r' 

83-5 

24-5 

I3i 

13-9 

.0.43 

ii.  i 

884.9 

128.7 

6.01 

294.5 

42.3 

3-47 

91.0 

26.8 

I3J 

14.0 

0--47 

ii.  i 

976.8 

140.8 

6.04 

325.4 

46.6 

3-49 

99-0 

29.1 

14 

14.0 

0.51 

ii.  i 

1070.6 

153-0 

6.07 

356.9 

51.0 

3-50 

. 

287-5 

84-5 

i6| 

14-9 

1.41 

ii.  i 

3836.1 

454-7 

6.74 

1226  .  7 

164.7 

3-81 

f 

64-5 

19.0 

nj 

II.  9 

0.39 

9-2 

499-0 

84.9 

5-13 

168.6 

28.3 

2.98 

H    12  J 
HI2 

71-5 

21.0 

i  if 

12.  0 

0.43 

9-2 

556.6 

93-7 

5-15 

188.2 

31-5 

3-00 

78.0 

22.9 

12 

12.0 

0.47 

9.2 

615.6 

102.6 

5.18 

208.1 

34-7 

3-01 

I 

161.0 

47-3 

134 

12-5 

0.94 

9.2 

1444-3 

214.0 

5-53 

477-0 

76-5 

3.18 

f 

49-0 

14-4 

9i 

IO.O 

0.36 

7.7 

263.5 

53-4 

4.28 

89.1 

17-9 

2.49 

H  10  \ 

54-0 
59-5 

15-9 
17.6 

10 
iof 

IO.O 
IO.O 

0.39 
0.43 

7-7 

7-7 

296.8 
331-9 

59-4 
65.6 

4-32 
4-35 

100.4 

112.  2 

20.  i 
22.3 

2.51 
2.53 

( 

123-5 

36.3 

114 

10.5 

0.86 

7-7 

790.4 

137-5 

4.67 

259-3 

49-5 

2.67 

f 

31-5 

9-2 

7f 

8.0 

0.31 

6.1 

105-7 

26.9 

3-40 

35.8 

8.9 

1.08 

H  8  \ 

34-5 

10.2 

8 

8.0 

0.31 

6.1 

121.  5 

30.4 

3.46 

41-  1 

10.3 

2.01 

39-0 

ii.  5 

8i 

8.0 

6.1 

139-5 

34-3 

3-48 

47.2 

II.  7 

2.03 

1 

90-5 

26.6 

94 

8.5 

0.78 

6.1 

385-3 

81.1 

3.8o 

125.1 

29.6 

2.17 

MATERIALS 
GIRDER  BEAMS 


i 

1   . 

|j 

"o  «5 

£ 

Neutral  axis  perpen- 
dicular to  web  at 

Neutral  axis 
coincident 
with  center 

>    ' 

^1 

^  "j 

IN 

w-C 

8   . 

rH   i/l 

center. 

line  of  web. 

u*s 

•&  s 

fc-3 

8.S 

Q>    O 

S  s 

I'2 

*a 

QS5 

a§ 

s& 

"Sg 

=1  3 

X'~ 

•"•S 

51 

8   .a 

1* 

g| 

g    .3 

|i 

la" 

1 

•s 
£ 

Is? 

H£ 

3J 
& 

!°! 

It 

rt  *> 

11 

P»l 

s  -a 

II 

—   00 

|1 

D 

t 

F 

I 

r 

i 

e 

/' 

r' 

n 

T 

30 

200.  o 

58.71 

0.750 

15.00 

9150.6 

12.48 

610.0 

630.2 

3-28 

II.  OO 

25.2 

3° 

180.0 

53-00 

0.690 

13-00 

8194.5 

12.43 

546.3 

433  3 

2.86 

9.00 

25.2 

28 

180.0 

52.86 

0.690 

14-35 

7264.7 

11.72 

518.9 

533-3 

3-18 

10.25 

23-4 

28 

165.0 

48.47 

0.660 

12.50 

6562.7 

11.64 

468.8 

371-9 

2.77 

8.50 

23-4 

26 

160.0 

46.91 

0.630 

13.60 

5620.8 

10.95 

432.4 

435-7 

3-05 

9-50 

21.6 

26 

150.0 

43-94 

0.630 

12.00 

5153.9 

10.83 

396.5 

3M.6 

2.68 

8.00 

21.6 

24 

140.0 

41.16 

0.600 

13-00 

4201.4 

IO.IO 

350.1 

346.9 

2.00 

?.oo 

20.  o 

24 

120.  0 

35.38 

0.530 

12.00 

3607.3 

IO.IO 

300.6 

249-4 

2.66 

.00 

20.3 

2O 

140.0 

41.19 

0.640 

12.50 

2934.7 

8.44 

293.5 

348.9 

2.91 

8.50 

15  7 

2O 

112.  0 

32.81 

0.550 

12.00 

2342.1 

8.45 

234-2 

239  3 

2.70 

8.00 

16.4 

18 

92.0 

27.12 

0.480 

II.SO 

1591.4 

7-66 

176.8 

182.6 

2.59 

7-50 

14.8 

15 

I40.O 

41.27 

0.800 

11-75 

1592.7 

6.21 

212.4 

331.0 

2.83 

7-75 

10.  1 

15 

104.0 

30.50 

0.600 

11.25 

I22O.I 

6.32 

162.7 

213.0 

2.64 

7-25 

ii.  i 

IS 

73-0 

21.49 

0.430 

10.50 

883-4 

6.41 

117.8 

123.2 

2.39 

6.50 

12.  1 

12 

70.0 

20.58 

0.460 

10.  OO 

538.8 

5-12 

89.8 

II4-7 

2.36 

6.00 

9.0 

12 

55.  o 

16.18 

0.370 

9.75 

432.0 

5-17 

72.0 

8l.l 

2.24 

6.00 

9-5 

10 

44-0 

12.95 

0.310 

9.00 

244.2 

4-34 

48.8 

57-3 

2.10 

5-50 

7-8 

9 

38.0 

11.22 

0.300 

8.50 

I7O.9 

3-90 

38.0 

44-1 

1.98 

5-25 

99 

8 

32.5 

9-54 

0.290 

8.00 

II4-4 

3-46 

28.6 

32.9 

1.86 

5.00 

6.0 

BEAMS 

30 

I2O.O 

35-3 

0.54 

10.5 

5239.0 

12.2 

349-0 

165.0 

2.2 

i 

6.50 

26.4 

28 

IOS.O 

30.9 

0.50 

10.  0 

4014.0 

II.  4 

287.0 

I3I.O 

.1 

i 

6.00 

24-7 

26 

90.0 

26.5 

0.46 

9-5 

2977-0 

10.6 

229.0 

IOI.O 

-9 

i 

5-50 

23.0 

24 

84.0 

24.8 

0.46 

9-3 

2382.0 

9.8 

198.0 

91.0 

-9 

5-25 

21.  0 

24 

73-0 

21.5 

0.39 

9-0 

2091.0 

99 

174.0 

74.0 

.9 

5-25 

21.3 

20 

82.0 

24.2 

0.57 

'  8.9 

I56o.O 

8.0 

156.0 

80.0 

.8 

5.00 

I7-I 

20 

72.0 

21.4 

0.43 

8.7 

1466.0 

8.3 

146.0 

76.0 

.9 

5.00 

17   I 

20 

69.0 

20.3 

0.52 

8.1 

1269.0 

7-9 

127.0 

Si.o 

.6 

4-50 

17-5 

2O 

64.0 

18.9 

0.45 

8.1 

1222.  O 

8.0 

122.0 

50.0 

.6 

4-50 

17-5 

2O 

59-0 

17.4 

0.38 

8.0 

II72.O 

8.2 

II7.0 

48.0 

•  7 

4-50 

17-5 

18 

59-0 

17.4 

0.50 

7-7 

883.0 

7-1 

98.0 

39-0 

.5 

4-25 

15-7 

18 

54-0 

15  9 

0.41 

7.6 

842.0 

7-3 

94-0 

38.0 

.5 

4-25 

15-7 

18 

52.0 

15-2 

0.38 

7.6 

825.0 

7-4 

92.0 

37-0 

.6 

4-25 

15-7 

18 

48.5 

14.2 

0.32 

7-5 

798.0 

7-5 

89.0 

36.0 

.6 

4.25 

15-7 

15 

71.0 

20.9 

0.52 

7-5 

796.0 

6.2 

106.0 

61.0 

•  7 

4-25 

ii.  7 

15 

64.0 

18.8 

0.60 

7-2 

665.0 

6.0 

89.0 

42.0 

•  5 

4.00 

12.3 

IS 

54-0 

15-9 

0.41 

7-0 

610.0 

6.2 

81.0 

38.0 

.6 

4.00 

12.3 

IS 

46.0 

13-5 

0.44 

6.8 

485.0 

6.0 

65.0 

25.0 

•  4 

3-75 

12.9 

15 

41.0 

12.0 

0.34 

6.7 

457-0 

6.2 

61.0 

24.0 

.4 

3-75 

12.9 

IS 

38.0 

ii.  3 

0.29 

6.7 

443-0 

6.3 

59-0 

23.0 

•  4 

3-75 

12.9 

12 

36.0 

10.6 

0.31 

6.3 

269.0 

5-0 

45-0 

21.0 

.4 

3-50 

9.9 

12 

32.0 

9-4 

0.33 

6.2 

228.0 

4-9 

38.0 

16.0 

•  3 

3-50 

10.2 

12 

28.5 

8.4 

0.25 

6.1 

216.0 

5-1 

36.0 

15-0 

•  3 

3-50 

10.3 

10 

28.5 

8.3 

0.39 

6.0 

134-0 

4-0 

27.0 

12.0 

.2 

3-25 

8.4 

IO 

23-5 

6.9 

0.25 

5-9 

123.0 

4-2 

25.0 

II.  0 

•  3 

3-25 

8.4 

9 

24.0 

7-0 

0.36 

5-6 

92.0 

3.6 

20.0 

9.o 

.1 

3.00 

7-5 

9 

20.  o 

6.0 

0.25 

5-4 

85.0 

3.8 

I9.O 

8.0 

.2 

3.00 

7-5 

8 

19-5 

S.8 

0.32 

5-3 

61.0 

3-2 

15-0 

7.0 

.1 

2.75 

6.6 

8 

17-5 

5.2 

0.25 

5-2 

57-0 

3-3 

14.0 

6.0 

.1 

2.  75 

6.6 

CHAPTER  II 
GRAPHICS 

Statics.  —  Statics  treats  of  forces  at  rest  and  therefore  in 
equilibrium,  hence  the  resultant  force  in  any  direction  and  the 
moments  of  the  forces  about  any  point  must  be  zero.  For 
coplanar  forces  the  condition  of  equilibrium  is  expressed  in  the 
following  equations: 

S  horizontal  forces  =  o, 
2  vertical  forces  =  o, 
2  moment  of  forces  about  any  point  =  o. 

Graphic  Statics.  —  Graphic  statics  relates  to  the  solution  of 
statical  problems  by  geometrical  constructions. 

Force.  —  Force  is  an  action  upon  a  body  tending  to  change 
its  state  of  rest  or  motion.  A  force  is  completely  known  when 
its  magnitude,  direction,  line  of  action  and  point  of  application 
are  known. 

Magnitude.  —  Forces  may  be  measured  by  any  unit  of  weight. 
It  is  most  convenient  in  this  work  to  use  the  pound.  The  magni- 
tude of  a  force  can  be  represented  graphically  by  the  length  of 
a  line,  the  length  being  drawn  to  a  previously  chosen  scale,  for 
instance,  if  the  scale  is  i  inch  to  1000  pounds  then  a  line  J-inch 
long  represents  a  force  of  500  pounds. 

Direction.  —  An  arrow  placed  on  the  line  is  used  to  indicate 
the  direction  of  the  force. 

Line  of  Action.  —  The  line  of  action  is  along  the  line  repre- 
senting the  force  and  it  is  along  this  line  that  the  force  tends  to 
produce  motion. 

Point  of  Application.  —  The  point  of  application  is  the  place 
assumed  as  a  point  where  the  force  acts  upon  the  body. 


GRAPHICS  21 

Coplanar  and  Noncoplanar  Forces.  —  Coplanar  forces  have 
their  lines  of  action  in  a  common  plane,  while  the  lines  of  action 
of  noncoplanar  forces  do  not  lie  in  the  same  plane.  Where  not 
otherwise  stated  coplanar  forces  are  to  be  understood. 

Concurrent  and  Nonconcurrent  Forces.  —  Concurrent  forces 
have  their  lines  of  action  intersect  in  a  point,  while  the  lines  of 
action  of  nonconcurrent  forces  fail  to  meet  in  a  common  point. 

Equilibrium.  —  A  number  of  forces  are  in  equilibrium  when 
they  produce  no  tendency  to  motion  in  the  body  upon  which 
they  act. 

Couple.  —  Two  equal  and  parallel  but  opposite  forces  con- 
stitute a  couple.  The  arm  of  the  couple  is  the  perpendicular 
distance  between  their  two  lines  of  action. 

Moment  of  a  Couple.  —  The  moment  of  a  couple  is  the  product 
of  either  force  by  the  arm  of  the  couple. 

Resultant.  —  The  resultant  of  a  system  of  forces  is  the  single 
force  or  simplest  system  that  would  produce  the  same  effect  as 
the  other  forces  and  could,  therefore,  replace  them. 

Equilibrant.  —  The  equilibrant  is  equal  in  magnitude  to  the 
resultant  but  opposite  to  it  in  direction.  It  is,  therefore,  the 
single  force  or  the  simplest  system  that  will  exactly  neutralize 
the  given  system  of  forces. 

Let  the  force  FI  acting  on  the  point  d  be  represented  in  direc- 
tion and  magnitude  by  the  line  ad, 
and  similarly  the  force  F2,  also  act- 
ing on  d,  be  represented,  by  dc,  then 
completing  the  parallelogram  and 
drawing  the  diagonal  db  gives  R,  the 
magnitude  and  line  of  action  of  the  F 

resultant.     Its  direction,  to  produce 

the  same  effect  as  the  two  forces  FI  and  F2,  must  be  as  indicated 
by  the  arrow. 

Force  Triangle.  —  The  triangle  dbc  would  have  given  the 
resultant  force  as  well  as  the  parallelogram.  Suppose  the  two 
known  forces  are  laid  off  so  that  the  arrows  run  in  the  same 


22  GRAPHICS  AND   STRUCTURAL  DESIGN 

direction,  that  is  either  clockwise  or  counterclockwise;  then  the 

tine  closing  the  triangle  with  its  ar- 
row following  around  in  the  same 
direction  as  the  other  two  arrows, 
here  counterclockwise,  will  be  equal 
to  the  resultant  but  opposite  in  di- 
rection; hence,  E  is  the  force  which 

would  hold  the  two  forces  FI  and  F2  in  equilibrium  and  is 
their  equilibrant. 

Force  Polygon. — If  more  than  three  forces  meet  at  a  point  and 
are  in  equilibrium  a  force  polygon  may  be  constructed  by  finding 
the  resultant  of  two  of  the  forces,  combining  this  resultant  with 
a  third  force  to  find  a  second  resultant,  and  so  on  until  all  the 
forces  have  been  considered  and  a  final  resultant  determined. 


FIG.  3.  FIG.  4. 

RI  is  the  resultant  of  F3  and  F4;  R%  is  the  resultant  of  RI  and 
F$,  and  F2  is  the  equilibrant  of  the  forces  FI,  F5,  F4  and  F3. 

It  is  evident  that  in  the  construction  the  drawing  of  the 
resultants  RI  and  R%  might  have  been  dispensed  with,  it  only 
being  necessary  to  lay  the  forces  off  so  that  the  arrows  would 
follow  in  the  same  direction.  It  follows  then  that  any  number 
of  concurrent  coplanar  forces  will  be  in  equilibrium  if  their  force 
polygon  closes  and  that  any  side  of  the  force  polygon  represents 
the  equilibrant  of  all  the  other  forces. 

Equilibrium  of  Coplanar,  Nonconcurrent  Forces.  —  In  the 
case  of  coplanar  nonconcurrent  forces  the  closing  of  the  force 
polygon  is  not  alone  sufficient  proof  of  equilibrium.  This  will  be 
seen  from  the  following  simple  example.  The  three  forces  FI,  F2 
and  ^3  are  equal  and  act  in  the  same  plane  making  an  angle  of 


GRAPHICS 


1 20  degrees  with  each  other.  It  is  evident  that,  being  an  equilat- 
eral triangle,  the  force  polygon  closes.  For  the  system  to  be  in 
equilibrium,  however,  the  equilibrant  of  F2  and  Fz  should  coin- 
cide with  E.  It  is  evident  that  the  equilibrant  of  the  system 
would  be  a  clockwise  or  negative  moment  FI  X  a.  Hence,  when 
a  system  of  nonconcurrent  forces  is  in  equilibrium  the  force 


FIG.  5. 


FIG.  6. 


polygon  must  close  and  the  sum  of  the  moments  of  the  forces 
must  be  zero.  It  follows  that  three  nonconcurrent  forces  can- 
not be  in  equilibrium  unless  the  forces  are  parallel  and  that  the 
resultant  of  the  group  of  nonconcurrent  forces  may  be  either 
a  single  force  or  a  couple. 

Components.  —  In  a  system  of  forces  having  a  resultant  each 
force  is  a  component  of  the  resultant.  Hence,  a  force  may  have 
any  number  of  components. 

Composition  of  Forces.  —  Composition  of  forces  consists  in 
finding  for  a  system  of  forces  an  equivalent  system  having  a 
smaller  number  of  forces.  The  simplest  case  of  composition  of 
forces  is  finding  a  single  force  replacing  a  system  of  forces. 

Resolution  of  Forces.  —  Resolution  of  forces  consists  in  finding 
for  a  system  of  forces  an  equivalent  system  having  a  greater 
number  of  forces.  An  illustration  of  this  is  where  a  given  force 
is  resolved  into  two  or  more  forces  or  components. 

EQUILIBRIUM  POLYGON 

A  system  of  forces  is  shown  in  Fig.  7.  Fig.  8  is  the  force 
polygon.  FI  and  F2  intersect  and  the  line  of  action  of  their 
resultant  must  pass  through  their  point  of  intersection  and  be 


GRAPHICS  AND   STRUCTURAL  DESIGN 


parallel  to  a,  the  resultant  of  FI  and  F2  in  the  force  polygon. 
Similarly  the  resultant  6  of  a  and  F3  must  pass  through  the  point 
of  intersection  of  a  and  F3  in  Fig.  7,  and  be  parallel  to  b  in  the 
force  polygon,  Fig.  8.  In  this  way  we  finally  reach  R,  the  result- 


FIG.  7. 

ant  of  F4,  with  b,  the  resultant  of  all  preceding  forces  FI,  F2  and 
F3.  In  Fig.  7,  R  is  located  by  the  fact  that  it  must  pass  through 
the  intersection  of  F4  and  b.  It  must  also  be  equal  and  parallel 
to  R  in  Fig.  8. 

Figure  TO  is  an  equilibrium  or  funicular  polygon.  For  the 
system  to  be  in  equilibrium  a  force  equal  and  opposite  to  R  would 
have  to  be  added  and  its  line  of  action  would  have  to  pass 
through  the  intersection  of  F4  and  b  in  Fig.  10. 

In  the  above  system  the  force  polygon,  Fig.  9,  closes  but  F5 
does  not  coincide  with  the  resultant  R  of  the  other  forces  FI,  F2, 


FIG.  9. 


FIG.  10. 


F3  and  F4,  in  the  equilibrium  polygon,  hence,  the  system  lacks 
equilibrium  by  the  clockwise  or  negative  moment  —  F5  X  d. 
The  system  is  in  equilibrium  for  translation.  The  resultant 


GRAPHICS  25 

of  the  system  is  —  F5  X  d.  The  equilibrant  of  the  system  is 
+  F5  X  d. 

The  method  just  given  applies  when  the  several  forces  can  be 
conveniently  made  to  intersect.  It  should  be  noticed  that  an 
infinite  number  of  equilibrium  polygons  are  possible  for  a  given 
system  of  forces. 

When  the  forces  do  not  intersect  conveniently  the  equilibrium 
polygon  is  drawn  as  follows: 

Take  any  point  O,  called  a  pole,  outside  of  the  forces  FI, 
F2  and  F3  and  connect  the  extremities  of  these  forces  with  the 
lines  called  rays,  a,  b,  c  and  d.  The  force  FI  will  be  hel'd  in  equi- 
librium by  the  two  rays  a  and  £,  acting  as  indicated  by  the 
arrows  inside  the  triangle  formed  by  the  force  FI  and  the  rays  a 


c-- 


FIG.  ii. 


FIG.  12. 


and  b.  In  the  same  way  b  and  c  hold  F2,  and  c  and  d  hold  F3  in 
equilibrium,  also  R,  the  resultant  of  FI,  F2  and  F3,  is  held  in 
equilibrium  by  the  rays  a  and  d.  Now  if  two  forces  represented  by 
the  rays  a  and  b  hold  FI  in  equilibrium,  then  in  the  equilibrium 
polygon  these  three  forces  must  intersect  in  a  point.  Take  any 
point  on  FI  in  Fig.  n,  and  through  it  draw  the  lines  called  strings 
parallel  respectively  to  b  and  a  in  Fig.  12.  F2  is  held  in  equi- 
librium by  the  rays  b  and  c\  therefore,  in  Fig.  n,  produce  b  until 
it  intersects  the  force  F2  and  through  .this  point  of  intersection 
draw  the  string  c  parallel  to  the  ray  c,  Fig.  12.  Through  the 
point  of  intersection  of  string  c  and  force  F3  in  Fig.  1 1  draw  the 
string  d  parallel  to  ray  d  in  Fig.  12  and  produce  it  until  it  cuts 
string  a.  Now  since  the  resultant  R,  Fig.  12,  is  held  in  equilib- 
rium by  the  rays  a  and  d,  these  three  forces  R,  a  and  d  must  in- 
tersect in  a  point  in  Fig.  1 1 .  The  intersection  of  a  and  d  locates  R 
in  Fig.  n,  its  magnitude  and  direction  being  given  by  Fig.  12. 


26  GRAPHICS  AND  STRUCTURAL  DESIGN 

Figure  12  is  the  force  polygon.  Fig.  n  is  the  equilibrium 
polygon.  The  difference  between  these  two  polygons  should  be 
carefully  noted.  The  force  polygon  gives  the  direction  and 
magnitude  of  the  forces  while  the  equilibrium  polygon  gives  the 
lines  of  action  and  the  direction  but  not  the  magnitude. 

GRAPHIC  MOMENTS 

Let  Fi,F2)  F3  and  F±  be  four  forces  constituting  a  system  whose 
bending  moment  is  desired  about  a  point  p.  Draw  the  force 


FIG.  13. 

polygon  (Fig.  14),  take  a  pole  0,  draw  the  several  rays  and  draw 
the  pole  distance  Os  perpendicular  to  the  resultant  R;  this  dis- 
tance is  H.  In  Fig.  13  draw  the  equilibrium  polygon  making  the 
strings  parallel  to  the  rays  in  Fig.  14.  Through  the  intersection 
of  strings  a  and  e  draw  the  resultant  R.  Through  the  point  p 
draw  y  parallel  to  R  and  limited  by  the  string  e  and  the  string  a 
produced.  The  bending  moment  of  R  about  the  point  p  is 
R  X  h.  The  triangles  123  and  4  0  5  are  similar,  hence 

R:H::y:h    or    Rxh  =  HXy. 

That  is,  the  bending  moment  of  any  system  of  coplanar  forces 
about  any  point  in  the  plane  is  the  product  of  the  pole  distance 
H,  and  the  line  y,  drawn  through  the  point  p,  parallel  to  the 
resultant  of  the  forces  R  and  limited  by  the  two  strings  e  and 
a  which  intersect  on  the  resultant. 


GRAPHICS 


27 


Moment  of  Parallel  Forces.  —  The  equilibrium  polygon  may 
be  used  to  determine  the  moment  of  any  or  all  of  a  system  of 
parallel  forces. 

The  bending  moment  on  the  beam  at  the  section  y-y  due  to 
the  forces  R^  and  FI  at  the  left  of  that  section  is  equal  to  1-2, 


-1-6 


± 


FIG.  15. 


FIG.  16. 


the  intercept  in  the  equilibrium  polygon  multiplied  by  the.  pole 
distance  H.     Expressed  algebraically  the  bending  at  y-y  equals 


M=(R1Xg)-  (Fl  X 


d) 


Now  triangles  135  and  680  are  similar,  hence  1-5  :g::  6-8:  H.  (2) 
Also  triangles  245  and  670  are  similar,  so  that  2-5 :  h ::  6-7 :  Z7.  (3) 

In  equation  (2),  6-8  =  RI,  therefore  RI  X  g  =  1-5  X  H. 
In  equation  (3),  6-7  =  FI,  therefore  FI  X  h  =  2-5  X  H,  and 

M  =  (1-5  X  #)  -  (2-5  X  H)  =  1-2  X  #. 

The  bending  moment  at  any  point  on  a  beam  due  to  a  system 
of  parallel  forces  is  equal  to  the  ordinate  of  the  equilibrium 
polygon,  cut  off  by  a  line  drawn  through  the  given  point  and 
parallel  to  the  resultant  of  all  the  forces  multiplied  by  the  pole 
distance.  By  bending  moment  on  any  beam  section  is  meant 
the  algebraic  sum  of  the  moments  to  the  left  of  that  section. 
The  intercept  in  the  equilibrium  polygon  is  a  distance  and  should 
be  measured  to  the  same  scale  as  that  to  which  the  beam  is  laid 
off.  The  pole  distance  is  a  force  and  should  be  measured  by  the 
scale  used  in  the  force  polygon. 


28  GRAPHICS  AND  STRUCTURAL  DESIGN 

USES  OF  FORCE  AND  EQUILIBRIUM  POLYGONS 

Reactions  of  a  Beam.  —  A  supported  beam  of  span  L  carries 
a  load  F  a  distance  a  from  the  left  support. 

Lay  off  F  in  Fig.  18,  take  the  pole  0  and  draw  the  rays  a  and  b. 
Through  any  point  2  on  line  of  action  of  load  F  in  Fig.  17  draw 
the  two  strings  parallel  to  the  rays  a  and  b.  Through  the  points 
i  and  3  where  the  rays  a  and  b  respectively  cut  the  lines  of  the 


I -c-------t 

"VjLx^     I. 


FIG.  17. 


FIG.  1  8. 


reactions  R\  and  R2  draw  the  string  c.  Now  in  Fig.  18  draw  the 
ray  c  through  the  pole  0  and  parallel  to  the  string  c  in  Fig.  17. 
This  ray  c  cuts  the  force  F  into  two  parts  representing  the  two 
reactions,  RI  being  held  in  equilibrium  by  the  rays  a  and  c  and  R2 
by  the  rays  c  and  b. 

Problem.  —  In  Fig.  19  find  the  reactions  when 
FI  =  10  tons  (20,000  Ibs.). 
F2  =  2  tons  (4000  Ibs.). 
^iw^ze'er.  —  RI  =  9300  Ibs.        R2  =  14,700  Ibs. 

Problem.  —  Draw  force  and  equilibrium  polygons  for  the 
beam  in  Fig.  20,  finding  reactions  RI  and  R%  and  the  maximum 
bending  moment.  Check  the  answers  by  algebraic  calcula- 
tions. 

Problem.  —  Draw  force  and  equilibrium  polygons  for  the  beam 
in  Fig.  21,  find  the  reactions  RI  and  RZ  and  the  maximum  bend- 
ing moment.  Check  the  answers  by  algebraic  calculations. 

In  treating  uniform  loads  graphically  it  is  necessary  to  divide 
the  load  into  small  sections  and  the  load  corresponding  to  each 
section  is  considered  at  its  center.  See  Figs.  22  and  23. 


GRAPHICS 


29 


l 


RI 


h 
l 


-T-  - 

FIG.  19. 


FIG.  20. 


iL      ^ 

|*--6L->j*-*!>)<--5 


—  20-- 

FIG.  21. 


Force  Polygon 

FIG.  23. 


K--8-W 


8^-»J 


FIG.  24. 


FIG.  25. 


•!      3"    i- 


-*«— ^ 


14- 


FIG.  27. 


FIG.  26. 


30  GRAPHICS  AND   STRUCTURAL  DESIGN 

Problem.  —  The  beam  shown  in  Fig.  24  carries  a  uniform 
load  of  1000  Ibs.  per  foot  of  length;  draw  the  shear  and  bending- 
moment  diagrams,  using  the  force  and  equilibrium  polygons. 

Determination  of  the  Center  of  Gravity.  —  Divide  Fig.  25 
into  convenient  regular  sections,  in  this  case  three  rectangles. 
Assume  forces  acting  through  the  centers  of  gravity  of  these 
rectangles  representing  their  areas.  Then,  by  means  of  force 
and  equilibrium  polygons  (Figs.  26  and  27),  locate  the  resultant 
or  equilibrant  of  these  three  forces.  The  location  of  the  vertical 
equilibrant  is  shown. 

Similar  treatment  of  these  areas  taken  as  acting  horizontally 
will  give  the  horizontal  equilibrant  and  the  center  of  gravity  of 
the  figure  will  lie  at  the  intersection  of  these  two  equilibrants. 

DEFLECTION  or  BEAMS 

For  beams  carrying  irregular  loadings  the  graphical  deter- 
mination of  deflections  is  convenient.  Books  on  mechanics  of 
materials  deduce  the  general  equation  of  the  elastic  curve  of 
d*  M 


If  the  equilibrium  polygon  representing  the  bending  moment 
upon  the  beam  is  known  the  curve  of  its  deflection  may  be  drawn 
in  the  following  way  : 

The  given  bending-moment  diagram  is  divided  into  sections 
and  the  areas  of  these  sections  are  represented  by  the  lines  A,  B, 
C,  etc.,  in  Figs.  28  (a)  and  (d).  In  the  funicular  polygon,  the  pole 
0  is  taken  a  distance  to  represent  El,  E  being  the  modulus  of 
elasticity  of  the  material  of  the  beam,  expressed  in  pounds  per 
square  inch,  and  /  the  moment  of  inertia  of  the  section,  in  inches4, 
and  referred  to  that  axis  of  the  beam  about  which  the  deflec- 
tion occurs.  Fig.  28  (c)  is  an  equilibrium  polygon  drawn  in  the 
usual  way  for  the  two  Figs.  28  (a)  and  28  (d).  The  intercepts  y 
in  the  diagram,  when  properly  scaled,  give  the  deflection  at  that 
point  on  the  beam.  To  determine  the  scale  if  i  inch  measured 
horizontally  in  Fig.  28  (a)  represents  s  inches  of  span,  i  inch  in 


GRAPHICS 


Fig.  28  (b)  equals  P  pounds,  H  the  pole  distance  in  Fig.  28  (b) 
measured  in  inches,  H1  the  pole  distance  of  Fig.  28  (d)  also 
measured  in  inches,  further  in  Fig.  28  (d)  i  inch  equals  k  square 
inches  of  the  bending  moment  area  of  Fig.  28  (a).  Then  i  inch 

of  intercept  in  Fig.  28  (c)  represents 


k-P 


H 


E.I 


inches. 


FIG.  28. 

One  square  inch  of  area  of  the  bending  moment  diagram, 
Fig.  28  (a),  represents  P  •  s2  -  H  (pound  inches2). 

The  correctness  of  this  method  may  be  shown  by  taking  a 
section  of  the  moment  curve  and  dividing  it  into  strips  having 
a  width  dx.  The  area  of  one  of  these  sections  is  M*  dx.  Com- 
paring the  similar  triangles  Oab  and  the  infinitesimal  triangle, 

M  •  dx     d2y        M       d*y 
we  have  — — —  =  — ^  or  — — -  =  — ^ .    This  is  the  general  equation 

Hi  •  JL          dX       Hi  •  L       dx 

of  the  elastic  curve,  as  well  as  the  equation  of  the  line  in 
Fig.  28  (c). 


32  GRAPHICS  AND   STRUCTURAL  DESIGN 

Continuous  Beams.  —  Another  use  of  the  curves  of  deflec- 
tion is  the  solution  of  restrained  or  continuous  beams  with 
irregular  loads  and  spans. 

The  following  is  a  modification  of  the  method  credited  to 
Dr.  Geo.  Wilson,  "Proceedings  of  the  Royal  Society,"  Vol.  62, 
Nov.,  1897. 


Force  and  Vector 
Polygons^ 


(ft) 


FIG.  29. 


The  method  consists  of  plotting  the  bending  moment  and 
deflection  curves  due  to  the  external  loads  assumed  as  carried  by 
the  outer  supports.  Similar  curves  are  then  drawn  for  the  same 
beam  excepting  that  now  the  actual  loads  are  removed  and  as- 
sumed reactions  are  applied  at  the  intermediate  supports,  the 
beam  being  held  as  before  at  the  outside  reactions.  If  the  beam 
is  level  the  deflections  caused  by  the  first  loading  must  be  equal 
and  be  opposite  to  those  produced  by  the  second  loading.  Equat- 


GRAPHICS  33 

ing  these  values  of  the  deflections  at  the  several  supports  gives 
sufficient  data  for  the  determination  of  the  reactions  and  when 
these  are  known,  a  revised  moment  diagram  for  the  actual  beam 
is  readily  made.  The  solution  is  simplified  by  using  the  same 
pole  distance  for  all  the  force  or  vector  polygons,  and  using  the 
same  scale  for  the  forces  in  all  diagrams.  An  example  will  show 
the  method  in  detail. 

A  continuous  beam,  Fig.  29  (a),  carries  three  loads  on  four  sup- 
ports. The  loads  are,  AB  =  1000  Ibs. ;  BC  =  800  Ibs. ;  and  CE  = 
1400  Ibs.  The  lengths  of  the  spans  beginning  with  that  at  A 
are  12  ft.,  10  ft.  and  8  ft.  The  loads  are  placed  as  shown. 

The  curve  of  deflection  Fig.  29  (e)  is  first  found  for  a  beam  sup- 
ported at  the  ends  and  carrying  the  three  loads,  AB,  BC  and  CD. 
The  force  polygon  Fig.  29  (c)  is  drawn,  from  which  the  bending 
moment  diagram  readily  follows  in  Fig.  29  (b).  This  polygon 
Fig.  29  (b)  is  then  divided  into  a  number  of  small  sections  of 
uniform  width,  similar  to  the  one  cross-hatched.  Since  they 
have  the  same  widths  their  areas  may  be  represented  by  their 
middle  ordinates,  shown  heavy  in  the  cross-hatched  section  and 
marked  5.  These  middle  ordinates  may  now  be  laid  off  in  the 
vertical  line  of  the  vector  polygon  Fig.  29  (d).  From  this  vector 
polygon  the  deflection  curve  Fig.  29  (e)  is  obtained  in  a  manner 
similar  to  that  used  for  the  bending-moment  diagram  Fig.  29  (b). 

Under  the  support  GF  the  deflection  in  Fig.  29  (e)  scales  51, 
while  under  FE  it  scales  40. 

Now  removing  the  loads  AB,  BC  and  CD,  assume  a  unit  load 
GF  applied  below  the  beam  and  acting  along  the  line  of  the 
reaction.  As  before  construct  a  force  polygon  Fig.  29  (f),  the 
equilibrium  polygon  of  bending-moment  diagram  Fig.  29  (g) 
and  finally  the  deflection  curve  Fig.  29  (i).  In  this  polygon  the 
deflections  for  the  unit  load  at  the  lines  of  the  reaction  scale  27 
and  19.  In  the  same  way  place  a  unit  load  at  EF  and  draw  the 
polygons  /,  k  and  m.  The  deflections  over  the  reactions  are  19 
and  1 8.  The  reactions  will  be  some  multiples  of  the  unit  loads 
applied.  These  unit  loads  may  be  any  convenient  one  as  i, 


34  GRAPHICS  AND   STRUCTURAL  DESIGN 

ioo,  500  or  1000  Ibs.,  depending  upon  the  beam  and  its  loading. 
If  we  call  the  reaction  GF,  p  times  the  unit  load  and  FE,  q 
times  the  unit  load,  then  the  deflections  due  to  GF  will  be  27  X  p 
and  19  X  p,  respectively,  while  those  due  to  FE  will  be  19  X  q 
and  1 8  X  q.  Since  the  deflections  at  the  points  of  support  due 
to  the  reactions  equal  the  deflections  due  to  the  loads  we  have 
in  this  problem,  under  GF 

(27  X  p)  +  (19  X  q)  =  51 
and  under  FE 

(19  X  p)  +  (18  Xq)=  40. 

Solving,  these  equations  give  p  =  1.26  and  q  =  0.92.  The  unit 
load  at  the  reactions  having  been  taken  at  1000  pounds,  the 
reactions  become  GF  =  1.26  X  1000  =  1260  Ibs.,  while  FE  = 
0.92  X  1000  =  920  Ibs. 

The  reaction  AG  may  be  found  by  taking  moments  about  the 
right  support  ED. 

Ar,      (ioooX24)+(8ooXi5)+(i40oX4)-(92QX8)-(i26oXi§) 

A(jr  =  ^— 

3° 
=  385  Ibs. 

Since  the  sum  of  the  reactions  equals  the  sum  of  the  loads  the 
reaction  ED  equals  (1000  +  800  +  1400)  —  (385  +  1260  +  920) 
=  635  Ibs. 

It  should  be  noted  that  the  pole  distances  were  all  equal  and 
the  same  scale  of  forces  was  used  in  all  force  polygons.  When  all 
the  forces  and  reactions  are  known  the  bending  moments  may  be 
found  either  graphically  by  constructing  the  force  and  equilib- 
rium polygons  as  usual,  or  the  moment  at  any  section  may  be 
found  by  computing  the  moment  of  the  forces  acting  on  one  side 
of  the  section. 


CHAPTER   III 
STRESSES   IN   STRUCTURES 

THE  application  of  force  and  equilibrium  polygons  to  the 
determination  of  stresses  in  structures  is  fairly  simple  espe- 
cially if  care  is  taken  in  properly  marking  the  structure  acted 
on  by  the  external^  forces  and  the  lines  of  the  diagram  as  drawn. 

In  the  simple  truss,  Fig.  31,  having  located  the  forces  at  the 
several  points,  place  a  letter  in  each  triangle  of  the  truss  and  a 
letter  between  each  pair  of  external  forces.  The  truss  being 


Upper  Chord 


FIG.  30. 


FIG.  31. 


FIG.  32. 


FIG.  33. 


FIG.  35. 


symmetrical  and  symmetrically  loaded  the  reactions  will  be  equal 
and  since  the  sum  of  the  vertical  forces  must  be  zero  each  re- 
action must  equal  one-half  the  total  load,  hence  the  reaction  is 
BC  +  CD  +  DP'  +  D'Cf  +  C'E' 

2 
35 


AB  = 


36  GRAPHICS  AND   STRUCTURAL  DESIGN 

To  assist  the  explanation,  numbers  have  been  placed  at  some 
of  the  points  but  this  is  not  usually  required.  Considering  the 
forces  acting  at  apex  i,  AB  and  BC  are  known  and  the  resultant 
of  these  is  held  in  equilibrium  by  forces  whose  lines  of  action  are 
CE  and  EA.  The  values  of  these  two  forces  can,  therefore,  be 
found  by  drawing  the  force  polygon  of  the  four  forces  acting  at 
point  i.  This  has  been  done  in  Fig.  32.  It  is  important  to 
place  the  directions  on  these  forces  and  then  indicate  these 
directions  as  acting  to  or  from  their  point  of  application.  This 
has  been  done  at  apex  i.  As  soon  as  an  arrow  is  placed  at  i 
on  CE  the  arrow  can  be  placed  on  CE  at  apex  2,  since  it  must  be 
opposite  in  direction  to  the  first  arrow.  Similarly,  the  arrow  on 
EA  can  be  placed  at  apex  3.  Now  considering  the  forces  acting 
at  apex  2,  EC  and  CD  are  known  and,  therefore,  the  other  two, 
DF  and  FE,  being  known  in  direction,  can  be  found  in  magnitude 
as  in  Fig.  33. 

Again  taking  the  forces  at  apex  3,  we  now  know  EF  and  EA 
from  which  FG  and  GA  can  be  found.  This  has  been  done  in 
Fig.  34.  In  this  manner,  force  polygons  could  be  drawn  for  all 
the  apices. 

Generally,  instead  of  drawing  these  force  polygons  separately 
as  just  done,  they  are  superimposed  upon  each  other  making  a 
diagram,  Fig.  35.  Where  the  truss  is  symmetrical  about  a  ver- 
tical axis  through  the  center  of  its  span  and  the  corresponding 
loads  on  the  two  sides  are  equal,  it  is  only  necessary  to  make 
the  portion  of  the  diagram  shown  in  full  lines.  The  other  half 
is  shown  dotted  and  by  rotation  about  AE  could  be  super- 
imposed upon  the  full-line  diagram.  In  this  case,  drawing  this 
dotted  diagram  is  unnecessary. 

TENSION  OR  COMPRESSION  IN  THE  MEMBER 

Where  the  forces  acting  upon  the  structure  are  indicated  as 
shown,  a  piece  having  the  arrows  < — >  shows  that  piece  to  be  in 
compression,  while  — »<—  indicates  tension. 


STRESSES   IN   STRUCTURES 


37 


NOTE.  —  In  analyzing  stresses  in  this  way  the  members  at  the  apices 
are  assumed  as  pin  connected  and  the  external  forces  of  the  pieces  are  indi- 
cated by  the  arrows  as  acting  on  the  pin.  That  is,  the  forces  acting  on  the 
pin  and  not  the  forces  of  the  pin  on  the  pieces  are  indicated  by  the  arrows. 

Frequently  trusses  carry  loads  attached  to  the  lower  chords. 
The  solution  graphically  follows  the  procedure  just  described. 

If  the  lower  apex  loads  BC,  CD,  etc.,  are  equal  the  reactions 
AB  and  AB'  will  be  equal  and  each  will  be  one-half  the  sum  of 
the  loads  on  the  lower  chord.  Taking  the  forces  at  point  i,  we 
can  lay  off  AB,  knowing  both  its  direction  and  magnitude,  then 
laying  off  BF  and  FA  parallel  to  the  respective  truss  members 


F-G 


FIG.  37. 


the  stresses  BF  and  FA  become  known  in  magnitude.  Now 
going  to  point  2,  since  AF  and  AG  will  be  parallel  lines  passing 
through  a  common  point  A  they  will  coincide  and  their  intercept 
on  FG  will  be  a  point  or  its  value  is  zero.  Hence  the  stress  in 
FG  due  to  loads  carried  on  the  lower  chord  is  zero. 

The  points  F  and  G  will  be  coincident  on  Fig.  37.  Now  going 
to  point  3,  GF,  FB  and  BC  are  known,  hence  CH  and  HG  are 
readily  drawn  and  their  magnitudes  determined. 

The  method  of  determining  the  direction  of  the  forces  is  the 
same  as  previously  explained  and,  as  before,  the  members  in  the 
upper  chord  are  shown  to  be  in  compression,  those  in  the  lower 
chord  in  tension. 

The  vertical  members,  excepting  LL' ',  will  be  in  compression 
while  the  diagonal  members  are  in  tension. 

Moving  Loads  Carried  under  Trusses.  —  Not  infrequently 
trolley  or  hoist  runway  tracks  are  carried  by  the  lower  chords 
of  trusses.  In  this  event  the  track  is  preferably  fastened  to  the 


GRAPHICS  AND   STRUCTURAL  DESIGN 


lower  chord  at  the  apices  of  the  triangles,  and  the  load  may  come 
upon  any  apex.  It  will  generally  be  easy  under  these  circum- 
stances to  make  a  diagram  for  the  load  under  each  apex  in  one- 
half  the  lower  chord,  and  by  comparing  the  diagrams,  which  are 
preferably  made  to  the  same  scale,  the  maximum  stress  in  any 
member  due  to  any  of  the  several  positions  of  the  load  is  readily 
found.  A  diagram  made  for  one  position  of  the  load  will  illus- 
trate the  method. 


F-G 

FIG.  40.  FIG.  42. 

The  load  is  assumed  at  apex  3.  The  reactions  AB  and  ABf  must 
first  be  found.  Draw  the  force  polygon,  Fig.  41,  and  the  equi- 
librium polygon,  Fig.  39.  When  the  ray  3,  in  Fig.  41,  is  drawn 
parallel  to  the  closing  side  of  the  equilibrium  polygon,  Fig.  39, 
the  load  BBf  is  divided  into  the  two  reactions. 

When  it  is  required  to  find  the  reactions  for  all  positions  of  the 
load,  the  work  can  be  abridged  by  laying  off  the  load  BBf  at  the 
left  reaction  and  taking  any  point  Q  in  the  other  reaction  and 
drawing  BQ  and  B'Q;  then  the  intercept  a-a  in  the  triangle  under 
the  load  will  be  the  left  reaction  for  that  position  of  the  load. 
If  BQ  and  B'Q  are  drawn  parallel  to  the  lower  chord  a-a  will 
be  the  left  reaction  and  a-b  the  right  reaction. 

NOTE.  —  The  student  should  prove  the  truth  of  this  last  diagram,  using 
both  algebraic  and  graphical  demonstrations. 


STRESSES   IN   STRUCTURES 


WIND  LOAD  ON  TRUSSES 


39 


The  wind  load  will  be  assumed  normal  to  the  left  side  of  the 
roof  and  both  ends  of  the  truss  will  be  fixed. 

The  reactions  AB  and  AB'  will  be  parallel  to  the  wind  loads 
BC,  CD,  DE  and  EF.  The  sum  of  the  reactions  will  equal  the 
sum  of  the  loads.  The  magnitude  of  the  reactions  can  be  found 
by  the  force  and  equilibrium  polygons,  Figs.  44  and  45.  In 
Fig.  44,  BBr  is  laid  off  parallel  to  the  wind  loads  and  equal  to 
their  sum.  The  resultant  normal  wind  pressure  on  the  roof 
FIG.  43- 


FIG.  46. 

equals  the  sum  of  the  wind  loads  BC,  CD,  DE,  EF  and  FB'  and 
will  act  coincident  with  the  force  DE.  Take  any  point  p  in  the 
line  of  this  resultant  and  through  it  draw  the  strings  parallel  to 
their  respective  rays  i  and  2  in  Fig.  44.  Then  in  Fig.  45  draw 
the  closing  line  of  the  equilibrium  polygon  and  in  Fig.  44  draw 
ray  3  parallel  to  string  3  in  Fig.  45.  BBf  will  then  be  divided 
into  two  parts  corresponding  to  the  two  reactions.  Having  found 
the  two  reactions  the  stress  diagram  follows  naturally  as  in 


40  GRAPHICS  AND   STRUCTURAL  DESIGN 

Fig.  46.  No  difficulty  is  experienced  until  apex  3  is  -  reached. 
Here  there  are  three  forces  known  completely  and  three  known 
in  direction  only.  It  is,  therefore,  impossible  to  determine  the 
magnitude  of  the  unknown  forces  unless  more  conditions  are 
assumed.  One  solution  of  the  problem  is  as  follows.  It  can  be 
shown  that  the  forces  acting  in  FL,  LM  and  MA  will  be  the 
same  whether  the  truss  is  left  as  shown  by  the  full  lines  JK  and 
KL,  or  if  these  are  removed  and  replaced  by  the  member  shown 
dotted,./!,. 

Taking  moments  about  point  5,  cutting  the  truss  as  shown  by 
the  line  and  equating  internal  and  external  moments  we  have 

2  external  moments  =  (force  FL)  X  a. 

As  the  other  two  forces  pass  through  point  5,  their  moments 
about  5  are  zero.  Hence,  the  force  in  FL  depends  only  upon  the 
external  moments  and  the  lever  arm  a,  and  is  consequently  inde- 
pendent of  the  truss  members  to  the  left  of  the  cut.  In  a  simi- 
lar way  stresses  in  LM  and  MA  can  be  shown  to  be  independent 
of  truss  members  to  the  left  of  the  cut.  Having  found  the  tem- 
porary forces  acting  at  3,  when  JK  is  removed  complete  the  dia- 
gram for  the  forces  at  5  and  6,  then  remove  the  dotted  member  JL 
and  replace  the  full  members  LK  and  KJ.  Now,  taking  the  forces 
acting  at  6,  EF  and  FL  are  known  completely;  therefore,  the 
magnitudes  of  EK  and  KL  can  be  found.  From  here  taking  the 
forces  at  point  7  and  then  returning  to  point  3  all  the  forces  act- 
ing in  the  left  half  of  the  truss  become  known.  The  diagram  can 
be  closed  by  finally  taking  the  forces  acting  at  the  right  reaction. 

The  stresses  due  to  the  wind  in  the  members  H'G'  to  L'M  in 
the  right  half  of  the  truss  will  be  seen  to  be  zero.  It  is  evident 
that  for  the  wind  acting  on  the  right  side  of  the  truss,  the  mem- 
bers in  the  right  half  will  be  stressed  the  same  as  similar  members 
in  the  left  half  when  the  wind  acts  as  shown  in  Fig.  46 ;  hence, 
redrawing  the  diagram  is  unnecessary. 

In  trusses  of  long  span  it  may  be  necessary  to  have  one  end 
free  to  allow  for  expansion;  this  may  be  done  by  allowing  the 


STRESSES   IN  STRUCTURES 


truss  to  rest  upon  a  plate  secured  to  the  wall.  The  plate  fastened 
to  the  lower  chord  of  the  truss  has  slotted  holes  which  permit 
the  truss  to  slide  upon  the  wall  plate  by  overcoming  the  friction 
between  the  plates.  The  friction  may  be  greatly  reduced  by 
using  rollers  under  the  truss.  The  following  example  illustrates 
the  method  of  determining  the  reactions  and  stresses  due  to  wind 
loading  when  the  truss  is  free  at  one  end.  The  plates  are  placed 
at  the  right-hand  end  of  the  truss.  The  coefficient  of  friction  is 
assumed  as  one- third;  hence  the  tangent  a  =  J  or  the  inclination 


FIG.  48. 


FIG.  47- 


FIG.  49. 

is  one  in  three.  The  line  of  action  of  the  right  reaction  is, 
therefore,  known.  The  first  part  of  the  problem  requires  the 
determination  of  the  magnitude  of  the  right  reaction  and  of  both 
the  magnitude  and  the  direction  of  the  left  reaction.  It  is 
known,  however,  that  the  left  reaction  must  pass  through  point  p 
of  the  truss.  -First  draw  the  force  polygon,  Fig.  48.  Through 
Bf  draw  a  line  parallel  to  the  right  reaction.  Take  any  pole  0 
and  draw  the  rays  i,  2,  3  and  4.  In  Fig.  47,  draw  the  cor- 
responding strings. 

Since  the  left  reaction  AB,  the  force  BC  and  ray  i  are  in  equi- 
librium they  must  intersect  in  a  common  point  in  the  equilibrium 
polygon;  this  must  be  point  p,  the  only  point  common  to  BC  and 
the  left  reaction  AB.  Hence,  through  p  draw  a  string  parallel 
to  ray  i.  From  Fig.  48  the  force  BC  is  held  in  equilibrium  by 
the  strings  i  and  2 ;  hence  in  Fig.  47  they  must  have  a  common 


GRAPHICS  AND  STRUCTURAL  DESIGN 


point.  Through  p  draw  also  the  string  2  parallel  to  ray  2.  Force 
CD  and  rays  2  and  3  are  in  equilibrium  and  must  pass  through  a 
common  point  in  the  equilibrium  polygon,  Fig.  47.  From  Fig.  48 
rays  4  and  5  hold  reaction  AB'  in  equilibrium  and  rays  i  and  5 
hold  reaction  AB  in  equilibrium;  hence  string  5  must  pass 
through  the  point  of  intersection  of  string  4  with  reaction  AB' 
and  it  must  also  pass  through  the  intersection  of  string  i  with 
reaction  AB  which  is  in  point  p.  It  is  noticed  that  string  i 
becomes  merely  a  point  in  Fig.  47.  This  work  would  have  been 
simplified  by  using  a  resultant  wind  pressure  equal  to  BB', 
acting  in  the  same  line  as  CD  in  Fig.  47. 


FIG.  50. 


FIG.  52. 

The  stress  diagram  is  shown  in  Fig.  49,  its  method  of  construc- 
tion differing  in  no  way  from  those  previously  described.  It 
should  be  noted  that  the  diagram  is  for  the  entire  truss,  the 
loading  being  unsymmetrical,  and  that  the  diagram  checks  if  the 
force  B'E'  when  drawn  through  the  points  Bf  and  E'  in  Fig.  49 
is  parallel  to  the  line  B'E'  of  the  truss,  Fig.  47. 

The  diagram  should  be  constructed  for  the  wind  on  the  free 
side  and  such  construction  is  shown  in  Figs.  50  to  52. 

When  rollers  are  used  at  the  free  end  the  reaction  at  that  end 
is  assumed  vertical. 

It  is  shown,  page  54,  how  the  stresses  may  be  obtained 
algebraically  by  the  method  of  moments  and  how  the  moments 


STRESSES  IN  STRUCTURES 


43 


due  to  external  loads  may  be  found  graphically.    A  combination 
of  these  two  methods  may  be  used  thus: 

Figure  56  is  the  force  polygon  for  the  external  loads,  with  pole 
0  and  the  rays  drawn.  From  this  the  equilibrium  polygon, 
Fig.  55,  has  been  drawn.  Now  the  bending  moment  at  any 
section  is  the  product  of  the  intercept  under  that  section  in  the 
equilibrium  polygon  multiplied  by  the  pole  distance  H.  Hence 
to  determine  the  stress  in  any  member  DF  cut  the  truss  at  Z-Z 
and  replace  the  cut  members  by  the  forces  that  would  have  to 
act  in  them  to  produce  equilibrium.  Now,  if  possible,  take 


FIG.  53. 


FIG.  54. 


FIG.  55. 


FIG.  56. 


moments  about  a  point  common  to  all  but  one  of  these  forces,  in 
this  case  point  i .  Let  a  be  the  lever  arm  of  the  force  DF  about 
point  i,  then  making  the  sum  of  the  internal  and  external  mo- 
ments zero  we  have 


(H  X  y)  +  (DF  X  a)  =  o, 


from  which 


The  Character  of  the  Stress,  whether  tension  or  compression, 
can  be  determined  as  follows: 

Calling  clockwise  rotation  positive  and  counterclockwise  neg- 
ative find  the  character  of  the  moment  of  the  external  forces 
H  X  y.  This  is  found  to  be  positive.  Since  the  sum  of  the 
internal  and  external  moments  about  a  point  is  zero  the  moment 


44 


GRAPHICS  AND   STRUCTURAL  DESIGN 


of  the  -internal  forces  must  be  of  opposite  character  to  the  mo- 
ment of  the  external  forces  hence  DF  X  a  is  negative  and  DF 
must  act  in  the«opposite  direction  to  that  assumed.  This  makes 
DF  in  compression. 


STRESSES   IN   STRUCTURES  45 

The  determination  of  the  dead-load  and  snow-load  stresses  in 
the  truss  of  a  bent  presents  no  greater  difficulties  than  that  of  a 
truss  carried  upon  brick  walls,  as  there  are  no  stresses  in  the  knee 
braces  due  to  these  loads.  Where  the  wind  load  is  carried  by  the 
columns  and  truss  forming  the  bent  the  magnitude  of  the  forces 
acting  in  the  truss  members  and  of  the  stresses  acting  in  the 
column  section  vary  greatly  with  the  distance  from  the  top  of 
the  column  to  the  foot  of  the  knee  brace  and  with  the  manner 
of  securing  the  column  to  the  foundation. 

The  wind  upon  the  truss  may  be  considered  as  either  horizontal 
or  normal  but  will  here  be  assumed  as  normal.  The  columns  will 
first  be  considered  as  hinged  top  and  bottom  and  then  as  fixed 
at  the  base  and  hinged  at  the  top.  In  this  latter  case  it  can  be 
shown  that  the  point  of  contraflexure  lies  between  the  base  of 
the  column  and  the  foot  of  the  knee  brace  and  between  one- 
half  and  five-eighths  of  this  distance  from  the  base  of  the 
column.  It  is  generally  assumed  as  one-half  for  convenience, 
and  this  is  sufficiently  accurate.  To  expedite  the  determi- 
nation of  the  stresses  the  forces  acting  upon  the  column  have 
been  transferred  to  the  truss  through  the  extra  truss  mem- 
bers added  temporarily  to  the  columns.  It  should  be  noted 
that  vertical  sections  can  be  taken  through  all  members  of 
the  truss  without  cutting  these  added  pieces,  hence  the  stresses 
in  these  members  will  be  independent  of  the  stresses  in  the 
added  pieces  and  the  diagram  drawn  with  their  assistance  will 
give  the  correct  stresses  in  the  permanent  members  of  the 
truss.  The  wind  load  has  been  assumed  as  20  Ibs.  per  sq.  ft., 
and  the  normal  pressure  has  been  estimated  by  the  formula 

p 

PN  =  —  X  A .     The  slope  is  i  in  4,  the  angle  being  approximately 

45 

27  degrees. 

rtXN 

PN  =  —  X  27  =  12  Ibs. 
45 


46 


GRAPHICS  AND  STRUCTURAL  DESIGN 


The  following  dimensions  have  also  been  used:  span  36  ft.  o  ins., 
panel  widths  16  ft.,  base  of  column  to  lower  chord  of  truss  14  ft., 
lower  chord  of  truss  to  foot  of  knee  brace  5  ft.  o  ins. 

In  Case  I  the  columns  are  hinged  both  top  and  bottom,  while 
in  Case  II  they  are  fixed  at  the  base  and  hinged  at  the  top.    The 


FIG.  60. 


FIG.  61. 


Scale 


fs* — ' — 4000^ — 


CASE  II 

COLUMN  BASE  FIXER, 
M"  COLUMN  TOP  -HI NGED 


lettering  has  been  made  identical  in  both  cases  and,  the  scales 
being  the  same,  the  influence  of  the  method  of  securing  the 
columns  upon  the  stresses  in  corresponding  members  can  be  seen 
at  a  glance  by  comparing  the  stresses  in  the  stress  diagrams, 
Figs.  59  and  62.  The  following  explanation  refers  to  Case  I,  but 
applies  also  to  Case  II.  It  is  first  necessary  to  estimate  the  apex 
wind  loads  on  the  sides  and  roof.  On  the  side  of  the  building 
the  wind  is  assumed  as  acting  at  the  base  of  the  column,  the 
foot  of  the  knee  brace  and  the  top  of  the  column.  The  normal 


STRESSES   IN   STRUCTURES  47 

wind  pressure  on  the  roof  has  a  resultant  WN  equal  to  the  sum 
of  apex  loads  EF,  FG,  GH  and  HBf,  and  acting  centrally  with 
this  side  of  the  roof  as  shown. 

Similarly,  the  horizontal  wind  pressure  WH  equals  the  total 
horizontal  wind  pressure  between  two  bents  and  acts  centrally 
upon  the  side  of  the  building.  Now  these  two  wind  forces  acting 
on  the  building  have  a  resultant  acting  through  their  point  of 
intersection  given  in  direction  and  magnitude  in  the  force 
polygon,  Fig.  58.  Taking  any  point  in  this  resultant  Rw  in 
Fig.  57,  an  equilibrium  polygon  can  be  drawn,  since  it  is  known 
that  the  resultants  AB  and  ABf  must  pass  through  the  column 
bases.  In  Case  II  these  resultants  pass  through  points  in  the 
columns  midway  between  the  base  of  the  column  and  the  foot 
of  the  knee  brace.  Having  drawn  the  strings  i,  2,  and  3  in 
Fig.  57,  draw  the  rays  i  and  2  in  Fig.  58,  and  their  point  of 
intersection  Q  will  be  the  desired  pole.  The  usual  assumption 
is  that  the  columns  share  the  horizontal  components  of  the  wind 
forces  equally.  In  Fig.  58  the  horizontal  component  of  the  wind 
forces  is  BS,  which  has  TU  drawn  perpendicular  to  it  from  its 
center.  If,  through  the  pole  Q,  ray  3  is  drawn  parallel  to  string 
3  in  the  equilibrium  polygon  in  Fig.  57  it  will  cut  TU  at  the  point 
of  intersection  of  the  resultants  AB  and  AB'  acting  at  the 
column  bases.  In  Case  II  these  resultants  act  on-  the  columns 
midway  between  the  bases  and  the  knee  braces.  Having  found 
the  resultant  wind  forces  acting  on  the  columns  the  drawing 
of  the  stress  diagram  presents  no  unusual  difficulties.  To 
transfer  the  forces  from  the  column  bases  to  the  truss  members 
C7,  •//,  JD,  B'l',  IfJf  and  J'B'  have  been  added. 

The  stresses  given  by  the  diagram  for  the  column  AI,  AT ', 
K'J'  and  KJ  will  not  be  the  correct  stresses  for  the  actual 
structure,  but  the  stresses  of  all  other  truss  members  including 
the  knee  braces  are  the  desired  ones.  In  the  drawings  of  the 
truss  the  members  in  compression  have  been  drawn  heavy. 

Maximum  Bending  due  to  Moving  Loads.  —  It  has  been 
shown,  page  26,  how  the  equilibrium  polygon  can  be  used  to 


GRAPHICS  AND  STRUCTURAL  DESIGN 


determine  the  bending  moments  upon  a  beam.  By  a  simple 
extension  of  the  principles  a  diagram  can  be  made  showing  the 
maximum  bending  that  will  occur  along  the  beam  as  a  system  of 
moving  loads  passes  over  the  span. 

The  loads  EC,  CD,  DE  and  EF  in  Fig.  63  are  a  constant  dis- 
tance apart  and  roll  across  the  girder  ab.  The  several  positions 
of  the  system  of  loads  relative  to  the  girder  can  either  be  shown 
by  redrawing  several  additional  positions  of  the  loads  or  by 
redrawing  the  girder,  moving  the  girder  under  the  loads.  The 


•*— H— -*• 


FIG.  64. 

latter  method,  being  the  easier,  has  been  used  and  two  additional 
J  positions  of  the  girder  c-d  and  e-f  have  been  drawn.  Fig.  64  is 
/  the  force  polygon  for  the  four  wheel-loads,  a  pole  0  was  chosen 
V  and  the  rays  i  to  5  inclusive  were  drawn. 

Draw  the  equilibrium  polygon  for  the  several  loads,  the  strings 
being  drawn  parallel  to  their  respective  rays.    Strings  i  and  2  will 
intersect  on  the  force  BC,  strings  2  and  3  will  intersect  on  force 
CD,  etc.    Now  draw  the  closing  line  of  the  first  equilibrium  poly- 
gon by  drawing  gh  through  the  points  of  intersection  of  the  ex- 
tended reactions  AB  and  AF  of  the  girder.      Then  for  the  first\ 
position  of  the  loading  upon  the  girder  the  bending  moment  at  any   J 
point  on  the  girder  will  be  the  product  of  the  intercept  measured  ^ 
in  feet  to  the  scale  of  the  span  by  the  pole  distance  H  measured  i 
by  the  scale  of  forces.     In  the  same  way  move  the  girder  to  the 
second  position  c-d  and  locate  the  closing  line  of  the  second  equi- 
librium polygon  in  the  line  io.    Proceed  in  this  way  until  the  loads 


J 


STRESSES  IN   STRUCTURES  49 

have  been  rolled  across  the  girder.  It  now  remains  to  compare  the 
bending  moments  at  definite  points  on  the  girder  for  the  several 
positions  of  the  loads.  Suppose  it  is  desired  to  know  the  maxi- 
mum bending  moments  at  p  intervals  across  the  girder,  one  would 
begin  by  comparing  the  intercepts  in  the  several  equilibrium 
polygons  a  horizontal  distance  p  from  the  left  end.  Several  of 
these  ordinatesjV,  kl  and  mn  have  been  drawn.  The  comparison 
may  be  assisted  by  drawing  a  curve  through  the  points  of  inter- 
section of  these  ordinates  with  the  closing  lines  of  their  respective 
equilibrium  polygons. 

This  is  shown  by  the  curves  RR  and  SS. 

The  intercepts  between  the  curves  RR  and  55  and  the  broken 
line  of  the  equilibrium  polygon  will  give  the  bending  moments 
exactly  only  at  the  vertexes  of  the  equilibrium  polygons;  there- 
fore, when  the  maximum  appears  to  be  between  vertexes,  the 
ordinate  should  be  checked  by  drawing  an  equilibrium  polygon 
with  the  loads  in  a  position  to  bring  a  vertex  at  the  desired  point. 

Diagram  of  Maximum  Live-load  Shears.  —  This  will  be  ex- 
plained by  a  loading  somewhat  resembling  the  usual  locomotive 
and  train  load. 

In  making  the  diagram  of  maximum  shears,  Fig.  66,  lay  off 
the  forces  BC,  CD,  etc.,  and  take  a  pole  distance  OB  equal  to 
the  girder  span.  Draw  the  rays  i  to  10  and  beginning  at  O 
draw  the  strings  i  to  10  parallel  to  their  respective  rays  and  in  ac- 
cordance with  the  usual  method,  i.e.,  CD  in  the  force  polygon  is 
held  in  equilibrium  by  rays  2  and  3;  hence,  in  the  equilibrium 
polygon,  strings  2  and  3  must  intersect  force  CD  in  a  common 
point.  The  side  of  the  equilibrium  polygon  under  the  uniform 
load  is  a  parabola  and  can  be  drawn  tangent  to  the  broken  line. 
To  understand  the  theory  of  the  diagram  draw  the  closing  line  OA , 
string  1 1 ,  of  the  equilibrium  polygon ;  this  is  also  ray  1 1  of  the  force 
polygon  and  divides  the  line  of  the  forces  B-H  into  the  reactions. 
B-A  is  the  left  reaction  when  the  load  BC  is  over  the  left  support 
with  the  other  loads  as  shown.  Suppose  the  loads  moved  to  the 
right  a  distance  a  or  the  same  relative  position  of  the  loads 


5° 


GRAPHICS  AND  STRUCTURAL  DESIGN 


and  girder  is  more  readily  obtained  by  moving  the  span  a  dis- 
tance a  to  the  left  as  designated  by  position  2.  The  closing 
line  in  this  case  is  string  12  and  the  reaction  is  MN.  As  the 
loads  travel  to  the  right  the  shears  under  load  BC  can  be  obtained 
by  measuring  from  point  B  the  distance  the  load  BC  is  to  the  right 
of  the  left  reaction  and  at  this  point  measuring  the  ordinate  from 
OB  to  the  broken  line  2,3-  -  to  10  of  the  equilibrium  polygon. 
This  of  course  must  be  measured  by  the  same  scale  as  that  to  which 
the  forces  are  laid  off.  When  the  usual  locomotive  wheel  loads  are 

FIG.  65. 


5pan-Positioa*« »J 

FIG.  66. 

preceded  by  a  much  lighter  pilot  wheel  there  may  be  a  question 
as  to  whether  the  greatest  shear  at  any  section  occurs  when  the 
pilot  wheel,  load  BC,  is  at  the  section  or  the  first  driver,  load 
CD,  is  there.  In  Fig.  66,  measure  the  span  marked  position  3, 
produce  string  2  until  it  cuts  PQ,  the  line  of  the  right  reaction. 
From  the  point  where  string  2  cuts  BH  draw  a  heavy  horizontal 
line  to  the  left.  To  find  the  maximum  shear  a  distance  b  from 
the  left  support  draw  the  intercept  c-dj  a  distance  b  from  the  line 
QP ;  this  will  be  the  shear  when  the  load  CD  is  at  the  distance  b 
from  the  left  support.  Compare  c-d  with  the  intercept  e-f  drawn 
between  the  broken  line  of  the  equilibrium  polygon  and  the 
horizontal  line  OB  and  at  the  distance  b  from  the  line  of  forces 


STRESSES  IN   STRUCTURES  51 

HB.  In  this  way  the  maximum  shear  may  be  found  for  any 
point. 

This  method  is  applicable  to  the  determination  of  the  maximum 
stresses  in  truss  members  as  well  as  in  plate  girders. 

The  following  illustrates  the  method  applied  to  the  determi- 
nation of  the  stresses  in  a  Pratt  truss. 


Example.  —  The  Pratt  truss,  Fig.  67,  spans  150  ft.  It  has 
6  panels  and  height  of  30  ft.  The  loading  is  Cooper's  E-6o;  see 
page  71,  under  Influence  diagrams,  for  this  loading. 

Find  the  stresses  in  BC,  be  and  diagonal  bC,  locating  the 
positions  of  loads  for  maximum  stress  and  shear  by  influence 
diagrams  and  using  equilibrium  diagrams  for  determining  the 
maximum  bending  moments  and  shears. 


CHAPTER   IV 
ALGEBRAIC   DETERMINATION    OF   STRESSES 

THE  conditions  of  equilibrium  used  in  the  determination  of 
stresses  are: 

1.  Sum  of  the  horizontal  forces  =  o. 

2.  Sum  of  the  vertical  forces  =  o. 

3.  Sum  of  the  moments  of  the  forces  about  any  point  =  o. 
The  stresses  in  the  members  cut  by  the  section  a-a  can  be 

replaced  by  forces  FI,  F2  and  F3,  equal  to  the  respective  stresses 


FIG.  69. 


in  these  members.  These  forces  FI,  F2  and  F3  hold  the  portion 
of  the  truss  to  the  left  of  the  section,  Fig.  68,  in  equilibrium  and 
the  stresses  in  the  members  may  be  determined  by  using  con- 
ditions i,  2  and  3. 

Referring  to  conditions  i  and  2,  they  should  be  understood  to 
mean  that  the  sum  of  the  components  along  any  line  must 
=  o,  and  the  sum  of  the  components  along  a  line  at  right  angles 
to  the  first  line  must  =  o. 

Frequently  the  work  may  be  abridged  by  using  the  more  gen- 
eral scheme.  Forces  and  components  acting  to  the  right  or  up- 
wards are  positive,  those  acting  to  the  left  or  down  are  negative. 
Clockwise  moments  are  positive;  counterclockwise  moments  are 

52 


ALGEBRAIC   DETERMINATION  OF   STRESSES  53 

negative.     To  find  the  stresses  in  CE  and  EA  in  Fig.  70,  cut 
the  truss  at  a-at  then,  by  condition  2, 

(AB  -  BC)  +  CE  sin  a  =  o, 

CE  =  -  (,4£  -  BC)  cosec  a, 

also  -CEcoso!  +  ,4£  =  o     or    AE  =  CEcos<x; 

here  AE  is  plus,  acts  to  the  right  and  produces  tension. 


In  a  similar  way  to  find  the  stress  in  DF  or  FE,  cut  the  truss  in 
section  b-b.     Considering  the  vertical  components  we  have 

(AB  -  BC  -  CD)  -  DF  .  sin  a  +  EF  sin  0  =  o. 
Now  considering  the  horizontal  components 

+  £.4  -  £F  cos  a  -  EF  cos  0  =  o. 

The  solution  of  these  equations  will  give 

the  stresses  DF  and  EF.     The  forces 

acting  at  a  point  may  also  be  treated 

by  this  method.    Considering  the  forces    BJ  /  \ 3   A    5       T|B' 

acting  at  point  i  in  Fig.  72,  and  using 

condition  2  we  have 

(AB  -BC)  -CEsma  =  o 
_CE=_(AB-BC). 
Sin  a 

The  force  CE  acts  down  towards  point  i.  ^45  —  CE  cos  a  =  o; 
;4E,  being  plus,  acts  towards  the  right.  The  direction  and  char- 
acter of  the  forces  can  be  checked  by  drawing  the  force  polygon 
for  the  forces  at  the  point. 


54 


GRAPHICS  AND   STRUCTURAL  DESIGN 


METHOD  or  MOMENTS 


The  stresses  in  framed  structures  can  be  determined  alge- 
braically by  placing  the  sum  of  the  external  and  internal  moments 
equal  to  zero. 

To  determine  the  stress  in  any  member  cut  the  truss  by  a 
section  passing  through  the  member  whose  stress  is  desired. 
Taking  the  portion  of  the  truss  to  the  left  of  the  section  replace 
the  members  by  forces  acting  to  the  right  of  the  section  as  shown 
in  Fig.  74.  Now  select  some  point  about  which  to  take  moments. 
If  possible,  take  a  point  through  which  all  the  cut  members  pass, 
excepting  the  member  whose  stress  is  desired.  In  the  example 


FIG.  73- 


given,  members  FG  and  GA  pass  through  point  3  so  that  moments 
will  be  taken  'about  point  3.  To  determine  the  direction  of  the 
force  acting  in  the  member  assume  that  it  acts  away  from  the  cut 
section  or  to  the  right  and  that  the  piece  is  in  tension.  If  the 
solution  gives  a  plus  value  the  assumption  was  correct;  if  the 
value  is  minus  the  piece  is  in  compression.  As  before  clockwise 
moments  are  positive,  while  counterclockwise  moments  are 
negative. 

Taking  moments  about  point  3  in  Figs.  73  and  74,  we  have 
Moments  of  external  forces  +  moments  of  internal  forces  =  o. 

[(AB  -  BC)  b]  -  [CD  X  (b-d)]  +  DFXe  =  o, 

-  [(AB  -  BC)  b]  +  [CD  X  (b  -  d)] 
e 


DF 


ALGEBRAIC   DETERMINATION  OF   STRESSES 


55 


The  quantity  (AB  -  BC)  b  will  be  much  greater  than  CD  (b  -  d) 
and  DF  will,  therefore,  be  negative,  and  consequently  the  piece 
DF  will  be  in  compression  instead  of  in  tension  as  assumed. 

To  determine  the  stress  in  FG  take  moments  about  point  i, 
then 

(CDXd)-(FGxf)=o, 


FG  =  + 


CDXd 

f 


The  value  of  FG,  being  plus,  agrees  with  the  assumption  and  the 
member  is  in  tension. 


E    2       6    C4     -10    6     -12  -12  -10         -6 


«/  Y3/-2/  \2   -I/  Vl     O/  \0    +17  VI  +2/  V2   +37  V3 


+3  /V  \+8    V  +11  V  +12  V  +11   V  +8     V   +3 


\/\    7\ 


FIG.  76.     H 


FIG.  77. 


In  dealing  with  bridge  trusses  with  parallel  chords  an  abridg- 
ment of  the  method  of  resolution  of  forces  is  possible. 

In  Fig.  76  it  should  be  noted  that  all  the  stresses  are  multiples 
of  one  of  the  sides  of  the  small  triangle  ETB,  thus  F-D  = 
3  •  T-B;  E-F  =  3  •  E-T  and  E-D  =  3  •  E-B.  E-D  is  the  ver- 
tical shear  in  panel  D,  E-C  that  in  panel  C,  and  E-B  that  in 
panel  B.  E-B  =  load  W. 


GRAPHICS  AND   STRUCTURAL  DESIGN 


Expressed  in  terms  of  the  apex  loads  W  and  the  angle  9,  E-T 
W  -  sec  B  and  T-B  =  W  •  tang  6,  hence: 


Member. 

Stress. 

Member. 

Stress. 

E-F  and  F-G 
G-H  and  H-I 
I-J  and  J-K 
K-L 

sWsecO 
2Wsec8 
i  Wsec6 
oWsecd 

F-D 
E-G 
H-C 
I-E 
B-J 

3  W  tang  6 
6W  tang  6 
SWtangd 
10  W  tang  6 
ii  W  tang  0 

K-E 

12  Wtang  0 

It  is  seen  that  the  coefficients  of  the  diagonals  are  the  coeffi- 
cients of  the  shears  in  their  panels,  thus  the  shear  in  panel  D 
is  3  •  W,  hence  3  is  the  coefficient  on  E-F  and  F-G,  being  —  for 
compression  members  and  +  for  tension  pieces.  In  the  same 
way  the  shear  in  panel  C  is  2  -  W,  hence  the  coefficient  is  2  for 
diagonals  G-H  and  H-I. 

The  next  point  to  be  noted  is  that  the  sum  of  the  coefficients 
of  the  members  cut  by  the  sections  a-a,  b-b  and  c-c  must  in  each 
case  be  zero.  Section  a-a  cuts  members  E-F  and  F-D\  the 
coefficients  then  are  —  3+3=0.  At  section  b-b  the  coefficients 
of  D-F  and  F-G  are  each  +  3.  Now  if  the  coefficient  of  E-G  is  x, 

then  +3  +  3+*  =  °> 

hence  x  =  —  6. 

In  this  way  all  the  coefficients  can  be  found  and  stresses  com- 
puted by  multiplying  these  coefficients  by  W  •  sec  6  for  diagonal 
members  or  by  W  •  tang  6  for  horizontal  chords. 

Live  Loads.  —  To  analyze  the  influence  of  live  loads  at  the 
several  apexes  of  the  lower  chord,  construct  Fig.  77.  With  the 
load  at  apex  3,  the  left  reaction  is  y  P,  the  right  reaction  y  P. 
The  coefficients  of  the  web  members  to  the  left  of  the  load,  apex  3, 
are  f  while  those  to  the  right  are  —  y. 

In  the  same  manner  from  Fig.  78,  which  is  the  stress  diagram 
for  the  load  P  at  apex  5,  it  is  seen  that  the  coefficients  of  the 
web  members  to  the  left  of  apex  5  are  T  while  to  the  right  the 
coefficients  are  —  y. 


ALGEBRAIC   DETERMINATION   OF   STRESSES 


57 


58  GRAPHICS  AND   STRUCTURAL  DESIGN 

It  will  be  seen  that  for  maximum  chord  stresses  the  truss  should 
be  fully  loaded,  while  for  minimum  chord  stresses  there  should  be 
no  live  load  on  the  truss.  In  this  type  of  truss  all  loads  on  upper 
or  lower  chords  produce  tension  in  the  lower  chord  members  and 
compression  in  the  upper  chord  members. 

The  maximum  web  stress  will  be  produced  when  the  longer 
segment  from  that  panel  to  a  pier  is  fully  loaded.  The  minimum 
web  stresses  will  be  created  when  the  shorter  segment  from  panel 
to  pier  is  fully  loaded. 

PRATT  TRUSS 

The  analysis  of  the  stresses  in  a  Pratt  truss  can  be  done  by  the 
method  of  coefficients  in  the  same  way  as  the  Warren  truss  just 
described.  If,  as  is  sometimes  done,  parts  of  the  dead  loads  are 
assumed  as  applied  at  the  apexes  of  the  upper  chord  the  only 
members  whose  stresses  will  be  affected  are  the  vertical  ones. 
To  find  the  shear  in  these  verticals  cut  the  truss  diagonally 
through  the  vertical  member  whose  stress  is  desired  and  equate 
the  shears.  (See  Fig.  83.) 

/-/  =  (C-D)  -  (C-B)  -  (B-A)  -  (D-E). 

Had  there  been  no  loads  upon  the  upper  chords,  as  in  Figs. 
81  and  82,  then 

/-/  =  (C-D)  -  (C-B)  -  (B-A). 

The  maximum  chord  stresses  are  equal  to  the  sum  of  the  dead- 
and  live-load  chord  stresses,  while  the  minimum  chord  stresses 
are  those  due  to  the  dead  load  only.  The  maximum  and  mini- 
mum web  stresses  are  found  by  adding  algebraically  the  corre- 
sponding live-  and  dead-load  stresses. 

Since  in  this  truss  the  diagonals  can  take  only  tension  the 
necessity  for  counter  diagonals  to  care  for  stresses  due  to  un- 
symmetrical  live  loading  should  be  noted.  The  stress  in  the 
counter  is  the  same  as  the  compression  would  have  been  in  the 
piece  for  which  it  acts. 


ALGEBRAIC   DETERMINATION   OF   STRESSES 


59 


w 
FIG.  81. 


FIG.  82. 


6000 


FIG.  84. 


6o 


GRAPHICS  AND  STRUCTURAL  DESIGN 


WARREN  TRUSS 

Problem.  —  Find  the  dead-load  stresses  by  the  method  of 
coefficients  in  the  following  deck  Warren  truss,  Fig.  84.  Span 
100  ft.  Load  at  apexes  on  the  upper  chord,  12,000  Ibs. 

The  secant  of  45  degrees  is  1.414.     Tangent  of  45  degrees  is  i. 

Shear  in  panel  C  =  3.5  X  W  =  42,00x5  Ibs. 

Shear  in  panel  D   =  42,000  —  12,000  =  30,000  Ibs. 

Shear  in  panel  E    =  42,000  —  12,000  —  12,000  =  18,000  Ibs. 

Shear  in  panel  F  =  42,000  —  12,000  —  12,000  —  12,000  =  6000  Ibs. 


Stress  in  EG  =  -  3%  W  sec0    =  -  3.5  X  12,000  X 

Stress  in  HI  =  +  *\  W  sec  6    =  +  2.5  X  12,000  X 

Stress  in  JK  =  —  i|  Wsec  6    =  —  1.5  X  12,000  X 

Stress  in  KL  =  +    i  W  sec  6    =  +  0.5  X  12,000  X 

Stress  in  HA  =  +  3.5  W  tang  6  =  +  3.5  X  1 2,000  X 

Stress  in  KA  =  +  7.5  Wt&ngd  =  +  7.5  X  12,000  X 


.414  =  —  59,400  Ibs. 
.414  =  +42, 400  Ibs. 
.414  =  —  25,500  Ibs. 
.414  =  +  8,500  Ibs. 

=  +  42,000  Ibs. 

=  -f-  90,000  Ibs. 


Stress  in  DI  and  EJ    = 

Stress  in  FL 

Stress  in  CG 

Stress  I'D  GB  = 


6  W  tang  6  = 

8  W  tang  6  = 

o 

6000  Ibs. 


-6X  12,000  Xi 
-  8  X  12,000  X  i 


—  72,000  Ibs. 

—  96,000  Ibs. 


CHAPTER   V 
INFLUENCE   DIAGRAMS 

WHEN  a  system  of  concentrated  loads  moves  across  a  girder  or 
a  trussed  bridge  the  maximum  moment  or  shear  at  a  section,  or 
stress  in  a  given  member,  will  be  produced  by  a  certain  position 
of  the  system  of  loads  relative  to  that  section  or  member. 

Influence  lines  are  used  to  determine  this  position  of  the  loads 
producing  maximum  moments,  shears  or  stresses. 

Influence  Diagrams.  —  An  influence  diagram  shows  the  varia- 
tion of  the  effect  at  any  particular  point,  or  in  any  particular 
member,  of  a  system  of  loads  moving  over  the  structure.  In- 
fluence diagrams  are  commonly  drawn  for  a  load  of  unity.  The 
moments,  shears,  or  stresses  for  any  system  of  loads  can  be  com- 
puted from  the  intercepts  in  this  diagram  by  multiplying  them 
by  the  given  loads. 

Influence  diagrams  to  find  the  position  of  loading  to  give  maxi- 
mum moment  at  a  given  point  in  a  beam  or  girder  or  at  a  given 
joint  on  the  loaded  chord  of  a  truss. 

In  Fig.  85, 

SPi  is  the  resultant  of  moving  loads  to  the  left  of  point  3. 
ZP2  is  the  resultant  of  moving  loads  to  the  right  of  point  3. 

To  construct  the  influence  diagram,  Fig.  86,  for  the  bending 
moment  at  3,  compute  the  bending  at  3  due  to  a  unit  load  at 
this  point 


and 


If  p  is  laid  off  on  the  left  reaction  and  (S  —  p)  upon  the  right 
reaction  and  their  extremities  are  joined  to  the  ends  of  the  hori- 

61 


62 


GRAPHICS  AND  STRUCTURAL  DESIGN 


zontal  line  ea  by  ba  and/e  then  the  vertical  intercept  cd  represents 
P  P  for,  let  this  intercept  cd  be  x  then  by  similar  triangles 

o 


eab  and  dac,  - 


—  —  —  :.  x  =  *  -  ^         and  similarly  for  the 
p  o  o 

triangles  dec  and  ae/. 

(S  -  p)p 

±  -  -^-^- 

S 


S-p     S 


or    x= 


FIG.  85. 


.A 


FIG.  86. 


To  find  the  moment  at  a  given  point  3,  due  to  a  system  of  moving 
loads,  multiply  the  intercept  under  each  load  by  that  load  and 
take  the  sum  of  these  products.  From  Fig.  86 

M  = 


Now  move  the  system  of  loads  a  small  distance  dx  to  the  left, 
allowing  no  load,  however,  to  pass  on  or  off  the  span  or  across 
the  given  point  3.  Then  for  the  new  position 

M  +  dM  =  ^Pl  (! 


Subtracting  the  preceding  equation  from  this  we  have 
dM  =  -  ZPi  dyi  +  2P2  dy2. 


INFLUENCE  DIAGRAMS  63 

Now  examining  for  maximum  bending  by  placing  this  equation 
=  o,  we  have 

2Pidyi  =  2P2dyt.  (i) 

By  similar  triangles 


from  which 


= 
dx          S  dx      S 


dy*          p 

Substituting  this  value  in  equation  (i) 

ZPi  (S  -  p)  =  SP2/>; 
hence 


from  which 

SPtS  =  2  (Pl  +  P2)  £ 
or 

Pi       S(P!+P2) 

S7~       5 

This  may  be  expressed  by  stating  that  the  maximum  bending 
will  occur  at  a  section  when  the  average  load  to  the  left  of  the 
section  equals  the  average  load  on  the  entire  span. 

The  influence  diagram  to  find  the  position  of  the  loading  to  give 
a  maximum  moment  at  a  given  joint  on  the  unloaded  chord  of  a 
truss,  having  either  parallel  or  inclined  chords,  may  be  found  in  a 
similar  way. 

Figures  87  and  88  are  drawn  for  joint  4  on  upper  chord. 

If  the  system  of  load$  is  moved  a  distance  dx  to  the  left  the 
rate  of  change  in  the  bending  moment  is 


=  -  ZPi  dy,  -  SP2  dy,  +  2P3  dys-  (i) 


To  be  a  maximum  —r-  =  o. 
dx 


I-  GRAPHICS  AND  STRUCTURAL  DESIGN 

But  by  the  geometry  of  the  construction 

dy\  _  S  —  p  .  dyz  _£         _,     dy2  _  dS  —  pc 
dx  =       S     '    dx  ~  S  ~dx  =        cS 


FIG.  88. 

These  values  substituted  in  equation  (i)  give  the  position  for 
maximum  moment  at  joint  4. 

d 

™  P 

MAXIMUM  SHEAR 

Reactions  and  Shears.  —  By  definition,  shear  is  the  algebraic 
sum  of  the  vertical  forces  to  the  left  of  the  section.  When  the 
unit  load  has  moved  a  distance  x  to  the  left  of  the  right  support, 

x  X  i 
Fig.  90,  the  shear  to  the  left  of  the  load  is  — - — ,  and  since  this 

value  under  the  load  is  reduced  by  unity  the  shear  to  the  right 

L  —  x 
of  the  load  is  — - —  X  i. 


INFLUENCE   DIAGRAMS  65 

The  shear  under  the  load  for  any  position  is  given  by  the  inter- 
cept in  the  triangle.     For  a  system  of  loads  the  shear  to  the  left 


FIG.  89. 


FIG.  90. 


of  the  first  load  is  the  sum  of  the  products  of  the  several  intercepts 
in  the  triangle  by  their  respective  loads.  In  the  case  illustrated, 
Fig.  91,  the  left  reaction  or  shear  to  the  left  of  load  i  is 

P*y*. 


FIG.  91. 

If  the  entire  system  of  loads  is  moved  to  the  left  a  distance  x, 
but  no  loads  enter  or  leave  the  span,  R  increases  an  amount 

making  the  reaction 


66 


GRAPHICS  AND  STRUCTURAL  DESIGN 

MAXIMUM  SHEAR  AT  ANY  POINT 


The  vertical  shear  at  a  section  a  distance  x  from  the  left 
reaction,  Fig.  92,  is  VXl  =  2Py. 

Under  the  load  PI  this  is  reduced  by  the  amount  PI  making 
the  shear  to  the  right  of  PI,  F2  =  S  Py  -  PI.  If  the  loads  are 
moved  to  the  left  until  P2  is  at  section  x  from  the  left  reaction  the 


FIG.  92. 


shears,  providing  the  same  loads  only  are  now  on  the  span,  will 

have  increased  an  amount  ZP  —  >  making 

JLt 


Comparing  VXl  and  VX2  it  is  seen  that  VXl  will  be  the  greater  so 
long  as 


or 


L 


Under  these  circumstances  the  shear  at  x  will  be  a  maximum  with 
p 

PI  at  x  when  —  -  exceeds  the  sum  of  all  the  loads  on  the  span 
0 

p        ^p 

divided  by  that  span.     If  -p  =  —  -  the  shears  will  be  the  same 

0          L 

P         ZP 

at  x  with  either  PI  or  P2  at  that  point,  while  if  -^  <  —  the 

0  X/ 

maximum  shear  will  occur  with  P2  at  x. 


INFLUENCE   DIAGRAMS 


67 


Had  another  load  P6  come  on  the  span,  Fig.  93,  the  increase 
in  shear,  after  the  load  PI  had  passed  x  and  advanced  a  distance 
b  to  the  left  of  it,  would  be 


Here  SPi_5  represents  the  sum  of  the  loads  PI  to  P*>  inclusive,  but 
not  the  load  P6  just  assumed  as  coming  on  the  span,  while 
SPi_6  is  the  sum  of  the  loads  from  PI  to  P&  inclusive.  The  dis- 


Fj 

C 

pa 

•>  C 

P3 

X 

P4 

1)  C 

P 

b  c 

p, 

")  r 

b 

1 

i 

; 

*     *• 

—  L  

FIG.  93. 

tance  c  can  only  range  in  value  from  zero  to  c  =  b,  hence  the 
increase  in  shear  will  be  somewhere  between 

2/^.5  -  -  P!      and      2Pi_e  -  -  PI 

Where  the  first  expression  is  negative  and  the  latter  positive 
both  positions  of  the  drivers  should  be  tried. 

Load  PI  at  section  x  will  give  a  maximum  shear  when 
P       y  IP 

and  load  PZ  will  give  a  maximum  shear  when 

Pi      2Pi-5 


In  the  case  of  a  uniform  load  the  vertical  shear  at  x  is  a  maxi- 
mum when  the  portion  of  the  span  to  the  right  of  x  is  fully  loaded, 
while  the  shear  will  be  a  minimum  when  the  portion  to  the  left 
of  x  is  fully  loaded. 

Position  of  a  sy stein  of  moving  loads  to  give  a  maximum  shear  in 
any  panel  of  a  truss  with  parallel  or  inclined  chords. 


68  GRAPHICS  AND   STRUCTURAL  DESIGN 

In  Figs.  94  and  95, 

2Pi  is  the  sum  of  the  loads  to  the  left  of  the  panel. 
2P2  is  the  sum  of  the  loads  on  the  panel,  here  panel  2-3. 
SPs  is  the  sum  of  the  loads  to  the  right  of  the  panel. 
m  is  the  number  of  panels  to  the  left  of  panel  2-3. 
n  is  the  number  of  panels  in  the  truss. 
p  is  the  length  of  a  panel. 

FIG.  94. 


FIG.  95. 

The  influence  lines  for  loads  of  unity  ab  and  cd  are  drawn  as 
for  shear  in  a  beam  or  girder.  When  point  3  is  reached  the  in- 
tercept in  the  triangle  abc  is 

/  _    _i_  *  X  (n  —  m  —  i) 
n 

Now  as  the  load  moves  across  panel  2-3  the  shear  is  reduced 
gradually  until  when  point  2  is  reached  the  unit  load  is  deducted 
making  the  shear 

T         W  ~~  Wl  Wl 

gh  = I  = 

n  n . 

The  influence  diagram  for  the  shear  in  panel  2-3  is  given  by  the 


INFLUENCE   DIAGRAMS  69 

broken  line  chea.     The  maximum  positive  shear  in  panel  2-3  is 

V  =  ^Pzyz  +  SP2}>2  -  ZPiyi. 

Moving  the  loads  a  small  distance  dx  to  the  left,  no  load  passing 
a  panel  point  or  end  reaction,  the  shear  becomes 

V  +  dV  =  2P3(J3  +  dy*)  +  2P2(j2  -  dyz)  -  2Pi(yi  -  dyj. 
By  subtraction 

dV  =  2P3dy3  -  2P2dy2  +  ZPidyi.  (i) 

Comparing  the  differential  triangles  with  the  similar  triangles 
cgh,  hegf  and  feat  we  have 

^i=_JL;      ^?  =  ^U      and      ^?  =  JL. 
d#          w/>       rfa;         w/>  dx      np 

dV 
Now  dividing  equation  (i)  by  dx  and  putting  —  =  o,  to  ex- 

amine for  a  maximum,  we  have 

^_0=SP  *    2Pl!L=j+2p  * 

<fo  w/>  w/>  w/> 

and 


n 

This  states  that  the  vertical  shear  in  any  panel  will  be  a  maximum 
when  the  load  in  that  panel  equals  the  average  panel  load  for  the 
span.  It  is  necessary  that  a  load  near  the  head  of  the  train  be 
at  the  panel  point  to  the  right  of  the  panel  in  which  the  shear  is 
sought. 

MAXIMUM  FLOOR-BEAM  REACTION 

Figure  96  shows  two  panels  p\  and  pz  with  floor  beams  a,  b  and 
c  and  accompanying  stringers.  To  find  the  maximum  floor- 
beam  reaction  at  b  construct  the  influence  diagram,  Fig.  96, 
making  its  ordinate  at  ef  equal  one.  Let  2Pi  be  the  sum  of  the 
loads  in  panel  pi  and  S/>2  the  same  for  panel  />2,  then  for  a  maxi- 

y  p       y  P    i   y  P 

mum  floor-beam  load  -  -  =  -  —  .    Fig.  97  is  the  influence 

pi          pi  +  pz 

line  for  the  bending  moment  at  b. 


7o 


GRAPHICS  AND  STRUCTURAL  DESIGN 


Ordinates  y  and  y\  similarly  located  in  Figs.  96  and  97  will  be 
to  each  other  as  the  corresponding  intercepts  ef  and  hi  or 

y 
yi 

hence 


a    OOQO&    OOOOO 


FIG.  96. 


FIG.  97. 


To  find  the  maximum  floor-beam  reaction  determine  the  maxi- 
mum bending  moment  on  a  beam  whose  span  is  /  =  pi  +  p2  at  a 
distance  pi  from  the  left  support  a,  and  multiply  it  by  the  sum  of 
the  panels  pi  and  pz  and  divide  this  by  the  product  of  pi  and  p% ; 
generally  pi  =  pz,  so  that  the  maximum  reaction  will  ordinarily 
equal  twice  the  maximum  bending  moment  divided  by  the  panel 
width. 

1     1    1 


k 

j  e  p 


FIG.  98. 


When  the  stresses  are  determined  by  calculation,  advantage  is 
taken  of  the  following  principles  in  computing. 

The  moment  due  to  the  loads  about  the  section  under  P5  in 
Fig.  98  is 

Mi  = 


d^ 


00 


•8 


3 


4-(D 


10    «»    M    N     M 


mNOloOlOO 
rn  IQ  «*  R  ?  « 
fj  ^  «l  •*  «  H 

" 


=s  2  5  S  3 

;    M     -*  oo     fO    •* 


-OOO 

.io>ot^ 


srO'p 


r;    *t    IH     M_ 
" 


~^- 

OMi-iONrOt^oo 
PO   q.  oq   M_  10  >q.  -q.   q» 


-^POO 

a"  8  B 


72  GRAPHICS  AND  STRUCTURAL  DESIGN 

If  the  moment  is  desired  under  P6,  P6  being  a  distance  c  from  P5, 
then 

M  2  =  Pi  Ol  +  C)  +  P2  (*2  +  C)  +  .    .    .    P5C. 

In  the  second  instance  if  SP  is  the  sum  of  all  the  loads  on  the 
span  to  the  left  of  P6,  then 

M2  =  Mi  +  SP  X-c. 

In  this  way,  starting  at  the  head  of  the  train,  the  moments  of  the 
loads  to  the  left  about  a  point  under  each  successive  load  are 
calculated  and  noted.  Having  placed  the  loads  in  the  position 
to  give  maximum  bending  at  the  point  to  be  investigated,  first 
find  the  reaction  to  the  left  of  RI  and  then  the  bending  moment 
at  the  point  desired. 

If  Ms  is  the  bending  due  to  loads  PI  to  P6  about  the  section 
under  P7  the  bending  moment  about  R2  of  loads  PI  to  P7,  inclu- 
sive, will  be 

Mi  =  M  3  +  2  (Pi  -  P7)  e 

and  the  reaction  is 


where  S  is  the  span. 

The  bending  moment  at  a  section  under  P5  then  is 
M  =  RJ  -  ML 

The  work  of  calculating  stresses  due  to  moving  locomotive  and 
train  loads  is  facilitated  by  the  use  of  the  following  table  which 
has  been  computed  in  the  manner  just  described  for  Cooper's 
E-6o  loading. 

In  this  moment  table  the  consecutive  wheel  loads  are  numbered 
from  the  left  to  the  right,  and  the  distances  given  between 
adjacent  wheels,  the  sum  of  the  distances  from  the  train  load 
to  each  wheel  and  the  sum  of  the  loads  from  load  18  to  and 
including  each  wheel  load  are  also  given.  In  the  body  of  the 
table  above  the  heavy  zigzag  line  are  given  the  moments  of  all 
loads  between  the  heavy  vertical  line  at  the  right  of  the  horizontal 
row  of  moments  and  any  wheel  load  to  the  left,  about  the  vertical 


INFLUENCE   DIAGRAMS  73 

line.  Thus  the  moment  of  loads  from  8  to  18  inclusive  about 
the  vertical  line  marking  the  beginning  of  the  uniform  train 
load  is  8,785,500  ft.  Ibs.  (Both  pounds  and  foot  pounds  are 
expressed  in  thousands.) 

Below  the  heavy  zigzag  line  are  given  the  moments  of  the  loads 
to  the  right  of  the  heavy  vertical  line  in  that  row;  thus  the 
moment  of  the  loads  9  to  15,  inclusive,  about  load  9  is  3,720,000 
ft.  Ibs.  Below  the  body  of  the  table  are  given  the  sum  of  the 
distances  from  wheel  i  to  the  several  wheel  loads,  and  also  the 
sum  of  the  loads  from  load  i  to  any  other  load  including  both 
this  latter  load  and  load  i. 

A  simple  example  will  illustrate  one  use  of  this  table.  In  a 
girder  whose  span  is  80  ft.  what  is  the  left  reaction  when  load  i 
is  over  that  support? 

Wheel  14  being  79  ft.  from  wheel  i  when  i  is  on  the  left  pier, 
14  will  be  i  ft.  from  the  right  pier.  The  moment  of  wheels  i 
to  14  about  wheel  i  is  14,400,000  ft.  Ibs.,  read  below  the  zigzag 
line. 

The  right  reaction  is  this  moment  divided  by  the  span  or 

-  =  180,000  Ibs.     The  left  reaction  is  the  sum  of  the 
oo 

loads  on  the  span  minus  the  right  reaction  or  348,000  —  180,000 
=  168,000.  The  reaction  can  also  be  determined  by  using  the 
moments  above  the  zigzag  line.  The  moment  of  the  loads  from 
i  to  14  is  13,092,000  ft.  Ibs.,  but  load  14  being  i  ft.  from  the 
right  pier  the  moment  of  the  loads  as  placed  upon  the  girder 
about  the  right  pier  is  13,092,000  +  the  sum  of  the  loads  from 
i  to  14  multiplied  by  i  ft. 

M  =  13,092,000  +  (348,000  X  i)  =  1 3, 440,000  ft.  Ibs. 
The  left  reaction  is 

*3.44o.°°°  =  xeS.ooolbs., 
80 

the  same  as  found  before. 

The  use  of  this  table  and  of  the  rules  just  derived  by  means 


74 


GRAPHICS  AND   STRUCTURAL  DESIGN 


of  the  influence  diagrams  will  be  illustrated  by  the  following 
problem. 

A  through  Pratt  truss,  Fig.  99,  carries  a  train  load  represented 
by  Cooper's  E-6o  loading.  Find  the  stresses  in  the  members 
of  the  second  panel  from  the  left  pier. 

To  find  the  maximum  stress  in  member  El  determine  the 
maximum  bending  at  apex  4.  The  criterion  for  maximum 
bending  at  a  point  in  a  truss  is  that  the  average  load  to  the  left 
of  the  point  shall  be  equal  to  or  less  than  the  average  load  on 
the  entire  span. 


1         A            7           A7 
150' 

FIG.  99. 


W  t»  »  f^  f*>  ' 


=  Average  panel  load  on  bridge. 
=  57,330  Ibs. 


Trying  load  7  at  apex  4: 
Panel  load  to  left  not 
including  load  7 

^P°  =  51,500  = 

Trying  load  8  at  apex  4: 

116,000 

— ^ =  58,000; 

To  find  reaction  at  the  left,  take  moments  of  loads  about  right 
support.  When  load  7  is  at  apex  4,  which  is  50  ft.  from  the  left 
pier,  load  2 1  will  be  the  last  one  on  the  bridge  and  it  will  be  3  ft. 
from  the  right  pier.  The  moment  about  the  right  pier  then  is 

M  =  24,064,000  +  (344,000  X  3)  =  25,096,000  ft.  lbs., 
from  which  the  reaction  at  the  left  is 

„        25,096,000 

p- **-          -  =  167, 100  lbs. 


INFLUENCE   DIAGRAMS  75 

The  bending  moment  under  apex  4  is 

,,      / 2 5, 006.000  X  SON  ..  „ 

M  =  {   °    v  — —  )  —  2,155,000  =  6, 2 10,330  ft.  Ibs. 

\  15°  / 

Here  2,155,000  is  the  bending  due  to  loads  i  to  6,  inclusive, 
about  load  7. 


The  stress  in  member  El  is 
6,210,330 


207,110  Ibs* 


Stress  in  number  HI.  To  determine  the  live  load  stresses  in 
HI,  first  find  the  maximum  shear  in  panel  3-4.  For  the  shear  in 
any  panel  to  be  a  maximum  the  load  in  that  panel  must  be  equal 
to  or  less  than  the  average  panel  load  on  the  bridge. 


Load  in  panel. 

Average  load  in  panel. 

Load  4  at  apex  4 

Lbs. 
50  ooo 

Lbs. 
•204.  000/6  —  *jo  670 

Load  5  at  apex  4 

70,000 

7Q4.  OOO/6=  ^0,670 

The  maximum  shear  in  panel  3-4  occurs  with  wheel  4  at  apex  4. 
To  find  the  reaction  under  these  conditions  when  load  19  is  4  ft. 
to  the  left  of  the  right  pier  we  have 

17,784,000  +  (304,000  X  4)  =  1  9,000,000  ft.  Ibs. 

T,       ,.  10,000,000 

Reaction  =  -  -  =  126,670  Ibs. 

150 

The  reaction  at  3  due  to  loads  in  panel  3-4  is 

D      moments  of  loads  1-4  about  4 

K.  =  -  • 

25 
,,        480,000 

R  =  —    -  -  =  19,200  Ibs. 
25 

Maximum  shear  panel  3-4  =  126,670  —  19,200  =  107,470  Ibs. 
If  tried  for  load  3  at  apex  4,  the  maximum  shear  will  be  found  to 
be  the  same. 
Using  this  shear  the  live-load  stress  in  HI  is  found  to  be 


HI  =  107,470  X  =  139,890  Ibs. 


CHAPTER  VI 
TENSION,    COMPRESSION   PIECES   AND   BEAMS 

DESIGNING  any  piece  of  a  structure  requires  the  determination 
,  of  the  resisting  forces  called  forth  in  it  to  balance  the  external 
forces  acting  upon  it.  In  the  case  of  a  purely  tension  piece  this 
requires  that  the  minimum  or  net  section  multiplied  by  the 
allowable  working  fiber  stress  shall  not  exceed  the  total  force 
acting  in  the  piece. 

Compression  pieces  whose  lengths  do  not  exceed  five  times  their 
least  diameter  can  be  designed  by  assuming  the  total  load  equal 
to  the  allowable  working  fiber  stress  in  compression  times  the 
area  of  the  minimum  section.  In  both  cases  care  should  be  taken 
to  have  the  load  distributed  over  the  section  as  improper  applica- 
tion of  the  external  forces  to  the  piece  may  result  in  introducing 
bending  or  injurious  local  concentrations  of  stress  in  it. 

BEAMS 

Upon  the  following  fundamental  conditions  of  static  equilib- 
rium determinations  of  the  required  forces  or  moments  acting 
on  or  in  the  piece  are  made.  This  applies  whether  the  deter- 
minations are  made  algebraically  or  graphically. 

The  sum  of  all  the  vertical  forces  =  o. 

The  sum  of  all  the  horizontal  forces  =  o. 

The  sum  of  the  moments  of  all  forces  about  any  point  =  o. 

Reactions.  —  In  a  beam  acted  on  by  several  vertical  forces 
the  sum  of  the  reactions  or  forces  at  the  supports  must  equal 
the  sum  of  the  loads,  and  the  algebraic  sum  of  the  moments 
of  all  forces  and  reactions  referred  to  any  point  must  be  zero. 
In  Fig.  100 

Pi  +  P2  +  P*  =  #1  +  #2.  (i) 

76 


TENSION,   COMPRESSION  PIECES   AND   BEAMS 


77 


Taking  moments  about  any  point  in  the  reaction  R2  we  have 

M  =  Piai  +  P2a2  +  P&s  =  RJ  =  o.  (2) 


The  solution  of  equation  (2)  gives  RI.  R2  will  be  given  by  sub- 
stitution in  equation  (i). 

Vertical  Shear.  —  The  vertical  shear  at  any  section  is  the 
algebraic  sum  of  all  the  external  forces  on  the  left  of  that  section. 
Thus  a  shear  diagram  of  the  beam  is  shown  by  Fig.  101. 

At  the  left  support  the  vertical  shear  equals  the  reaction  RI, 
and  this  value  of  the  shear  continues  toward  the  right  until 
under  the  load  PZ  the  shear  becomes  R\  —  Pa;  similarly,  under 
P2  the  vertical  shear  is  RI  --  (P3  -f  P2)  and  under  PI  the  shear 
!  -  (P8  +  P2  +  Pi)  =  -  #2. 


s 


1 

1 

4 

u   c    , 

ft 

I 

2. 

i* 
i 

^i 

FIG.  10 

FIG.  101. 

Bending  Moment.  —  The  bending  moment  at  any  section  is 
the  algebraic  sum  of  the  moments  of  the  external  forces  on  the 
left  of  that  section  referred  to  a  point  in  that  section.  In  the 
beam,  Fig.  102,  the  bending  at  the  section  ab  is 

M  =  Ric  -  Pze  -  P2d. 

Where  rolled  beams  are  used  generally  only  the  maximum  bend- 
ing is  required;  this  will  occur  where  the  vertical  shear  passes 
through  zero.  (See  any  book  on  Mechanics  of  Materials  for 
the  proof  of  this.)  In  the  case  illustrated  in  Fig.  101,  the  maxi- 
mum bending  occurs  under  load  P2.  Having  found  the  maximum 
bending  moment,  if  the  beam  is  supported  laterally,  the  proper 


78  GRAPHICS  AND   STRUCTURAL  DESIGN 

section  can  usually  be  selected  from  a  manufacturer's  handbook. 
In  building  construction  beams  will  commonly  be  supported 
laterally  by  flooring,  roofing  or  bracing. 

The  external  bending  moment  induces  an  equal  and  opposite 
moment  in  the  material  of  the  beam  called  a  resisting  moment. 
This  is  expressed  by  equation 


where 
M  =  the  external  bending  moment  at  a  given  section. 

1—  =  the  resisting  moment  at  the  same  section. 

/  =  the  extreme  unit  fiber  stress,  generally  pounds  per  square 

inch. 

/  =  the  moment  of  inertia  of  the  beam  section,  in  inches. 
e  =  the  distance  from  the  neutral  axis  to  the  extreme  fibers 

in  inches. 

The  expression  -  depends  entirely  upon  the  form  of  the  beam 
e 

section  and  is  sometimes  called  the  section  modulus. 

An  example  will  illustrate  the  selection  of  a  floor  beam.  A 
floor  beam  with  a  span  of  12  ft.  is  to  carry  a  uniform  load  of 
14,400  Ibs.  Select  a  suitable  beam  allowing  a  working  fiber 
stress  of  16,000  Ibs.  per  sq.  in. 

The  maximum  bending  for  a  supported  beam  with  a  uniform 

Wl 

load  is  M  =  —  •  ,  where  W  is  the  total  uniform  load  in  pounds 
8 

and  /  is  the  length  of  the  span  in  inches. 

Substituting  the  given  values  in  this  formula  we  have 

1/r      14,400  X  (12  X  12)  .     ,, 

M  =  -       -  j  -  L  =  259,200  in.  Ibs. 
8 

The  handbooks  usually  tabulate  the  values  of  -  of  their  sections. 

C/ 

*  For  the  derivation  of  this  formula  see  any  book  on  "  Mechanics  of  Materials." 


TENSION,   COMPRESSION  PIECES   AND   BEAMS 


79 


By  transposition  of  the  previously  given  equation  -  =  — ,  then 

by  substitution 

7  _  259,200  _    , 
e       16,000 

The  lightest  weight  standard  I  beam  providing  a  sufficient  sec- 
tion modulus  is  a  Q-in.  I  beam,  weighing  21  Ibs.  per  ft.  Its 
section  modulus  is  18.9. 

Standard  Framing 


^"Clearance 


Clearance 


FIG.  103. 


Not  infrequently  the  beam  must  not  only  be  amply  strong 
but  it  must  not  deflect  excessively  under  load.  When  plastered 
ceilings  are  carried  under  the  beams  this  deflection  is  limited  to 
-§\-§  of  the  span.  The  formula  for  the  deflection  at  the  middle 
of  a  supported  beam  carrying  a  uniform  load  is 
WP  , 


A  = 


3»4  El 


or 


384EXA 


For  steel  E  =  30,000,000  Ibs.  per  sq.  in. 

If  the  deflection  is  A  =  ^  =  IM  =Hk 

360       360      5 

the  limiting  inertia  then  will  be 
/  = 


=     5  X  14,400  X  I443     =    6 
384  E  X  A      384  X  30,000,000  X  f  "        ''" 


It  is,  therefore,  evident  that  the  9-in.  I  beam  weighing  21  Ibs. 
per  ft.  demanded  for  strength  will  be  amply  stiff,  its  inertia 
being  84.9  or  almost  twice  the  inertia  required  for  stiffness. 


8o 


GRAPHICS  AND   STRUCTURAL  DESIGN 


In  framing  beams  into  girders,  beams  or  columns  each  manu- 
facturer has  a  standard  framing.  This  framing  is  designed  for 
the  shortest  span  and  consequently  the  greatest  load  for  which 
the  beam  is  likely  to  be  used. 


FIG.  104. 

The  standard  framing,  Fig.  104,  should  not  be  used  for  spans 
less  than  the  following: 


I. 

Lb.        Span, 
ft. 

7. 

Lb.         Span, 

ft. 

/. 

Lb.     Span, 
ft. 

7. 

Lb.       Span, 
ft. 

24 

80.0-22.0 

15 

8o.o-2O.o 

10 

25.0-9.0 

6 

12.25-6.0 

20 

8o.O-22.O 

15 

60.0-15.5 

9 

21  .O-7.O 

5 

9-75-4-0 

20 

65.0-18.0 

15 

42.O-II.O 

8 

18.0-5.5 

4 

7-5  -3-o 

18 

55.0-14.0 

12 

40.0-11.5 

'  7 

15.0-4.0 

3 

5-5  -2.0 

12 

•31    e—   n    o 

All  rivets  in  standard  framing  are  J  in.  in  diameter. 
As  this  goes  to  press  the  American  Bridge  Company,  in  their 
specifications  for  steel  structures,  publishes  a  revised  standard 


TENSION,    COMPRESSION  PIECES  AND  BEAMS  8 1 

for  framing,  in  which  all  beams  use  4  in.  X  4-in.  angles  from  27 
ins.  to  12  ins.  inclusive;  smaller  beams  use  6  in.  X  4-in.  angles. 

When  work  is  being  detailed  for  production  in  a  particular 
shop  the  standards  of  that  shop  should  be  ascertained  and 
adhered  to. 

It  should  be  noted  in  Fig.  103  that  to  facilitate  erection  the 
connecting  angles  extend  f  in.  beyond  the  end  of  the  web  of 
the  beam  and  that  the  distance  back  to  back  of  the  connecting 
angles  is  |  in.  less  than  the  space  into  which  the  beam  is  fitted. 
When  drawings  are  being  made  for  a  shop  having  standard 
framing  the  beam  sketch  may  be  like  Fig.  103,  no  dimensions 
being  placed  on  the  standard  rivet  spaces. 

The  advantages  of  the  standard  framing  are: 

It  simplifies  shop  work,  enabling  a  large  number  of  angles  to 
be  made  at  one  time.  The  punching  of  the  rivet  holes  both  in 
beams  and  angles  can  be  done  by  means  of  multiple  punches,  a 
group  of  holes  being  made  at  one  stroke  of  the  punch.  The 
drawing-room  work  is  also  reduced.  The  total  saving  in  time 
more  than  counterbalances  any  probable  waste  of  material.  A 
criticism  of  the  method  used  for  designing  these  connections  is 
that  there  is  twisting  introduced  into  the  rivet  groups  increasing 
the  shear  on  the  rivets.  Tests  made  on  some  full-sized  beams 
with  standard  framing  seemed  to  justify  the  usual  practice  as 
being  ample  notwithstanding  the  above  theoretical  criticism. 

When,  as  is  the  case  for  very  short  spans  or  for  beams  with 
thin  webs,  it  is  necessary  to  design  special  connections  the  cal- 
culations should  be  based  upon  the  shearing  and  bearing  value 
of  the  material;  see  page  84. 

Riveting. — There  are  two  types  of  rivet  heads:  "button 
heads,"  which  approximate  hemispherical,  and  " countersunk," 
which  are  truncated  cones.  The  button-headed  rivets  should 
always  be  used  where  clearances  will  permit.  Button  heads  can 
be  flattened  slightly  where  additional  clearance  is  needed  and 
this  requires  less  work  than  countersunk  rivets.  The  following 
illustrations,  Figs.  105  and  106,  give  the  dimensions  of  various 


82 


GRAPHICS   AND   STRUCTURAL  DESIGN 


.,*  '<• 


IT 
f-. 

I 


sap.IS 


© 

0 


c 


c 


TENSION,   COMPRESSION   PIECES  AND   BEAMS 


rivet  sizes  and  symbols  for  shop  and  field  driving,  and  are  taken 
from  the  American  Bridge  Company's  Book  of  Standards. 

Rivets  are  designated  by  their  nominal  diameter  D  and  the 
length  under  the  head  before  driving.  This  length,  Figs.  107 
and  1 08,  is  made  up  of  the  grip,  the  distance  through  the  material 
held  together  plus  the  length  called  the  upset  or  the  stock 
required  to  form  the  head  and  bring  the  body  of  the  rivet  up 
to  the  size  of  the  rivet  hole.  It  is  necessary  for  the  rivet  holes 
to  be  larger  than  the  rivet  stock,  otherwise  there  would  be 


h*— Grip-- 


•Upset — H 


FIG.  107. 


difnculty  in  placing  the  rivets  in  their  proper  rivet  holes  and  the 
stock  would  be  cold  before  they  could  be  driven.  The  holes  are 
usually  -^Q  in.  larger  in  diameter  than  the  diameter  of  the  rivets. 
The  number  of  rivets  required  in  any  joint  depends  upon  the 
forces  being  transmitted  by  the  rivets,  and  upon  the  rivet  values 
in  shear  and  bearing.  The  value  in  shear  may  be  that  due  to 
either  single  or  double  shear  as  the  rivet  tends  to  fail  by  shearing 
in  one  or  two  cross  sections.  The  bearing  value  of  the  rivet  is 
the  rivet  diameter  times  the  lesser  thickness  of  the  materials 
transmitting  the  forces  to  the  rivets  times  the  allowable  unit 
bearing  value  of  the  rivet  material.  The  following  tables  give 
the  shearing  and  bearing  values  of  the  usual  rivet  sizes. 


GRAPHICS  AND   STRUCTURAL  DESIGN 


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^ 

TENSION,   COMPRESSION   PIECES   AND   BEAMS  85 

A  few  examples  will  illustrate  the  application  of  the  above 
principles. 

fc  =  the  unit  bearing  value,  pounds  per  square  inch. 

/,  =  the  unit  shearing  value,  pounds  per  square  inch. 
Evidently  the  rivets  in  Fig.   109  would  tend  to  shear  in  but 
one  place,  along  ab,  and  are,  therefore,  in  single  shear. 

Shearing  value  =  -  -  X  /..         Bearing  value  =  d  X  /i  X  fc. 
4 

\t\         x~N /^"N  i 


a 

*  1                             * 

b 

c 

,    .'A;'".  "  .    ,f 

,/ 

] 


O      O 


OQ 


*!<  £2  Rivet-diam.=d.  ^<  tg  Rivet-diam.=d. 

FIG.  109.  FIG.  no. 

In  Fig.  no  the  rivets  are  in  double  shear  as  they  tend  to  fail 
along  the  sections  ab  and  cd. 

Shearing  value  =  --•/,.        Bearing  value  =  d  X  k  Xfc- 

2 

The  following  numerical  example  will  illustrate  the  design  of  a 
joint  like  this  one. 

Example.  —  A  joint  of  the  type  shown  in  Fig.  no  is  to  trans- 
mit 15,000  Ibs.;  /i  =  |  in.,  /2  =  f  in.,  rivets  f  in.  in  diameter, 
the  allowable  shearing  stress  10,000  Ibs.  per  sq.  in.,  allowable 
bearing  stress  20,000  Ibs.  per  sq.  in.  The  rivet  is  in  double 
shear  while  the  minimum  bearing  value  will  be  that  of  the  rivet 
against  the  |-in.  plate  as  in  the  other  direction  two  plates  each 
f  in.  thick  act.  From  the  tables  the  rivet  value  of  f-in. 
rivets  in  double  shear  at  10,000  Ibs.  is  4420  X  2  =  8840  Ibs. 
The  rivet  value  in  bearing  on  a  ^-in.  plate  is  7500  Ibs.  As  the 
joint  is  to  transmit  15,000  Ibs.  the  number  of  rivets  is  found  by 
dividing  15,000  Ibs.  by  the  smaller  of  the  two  rivet  values,  i.e., 
15,000 


=  2. 
7500 


86 


GRAPHICS   AND   STRUCTURAL   DESIGN 


Riveted  joints  should  be  designed  to  avoid  eccentric  stress  if 
possible.  Where  this  is  unavoidable  such  eccentricity  should  be 
reduced  to  a  minimum  and  the  resultant  forces  acting  upon  the 
individual  rivets  be  determined.  An  instance  of  such  a  joint, 
frequently  met  with,  is  the  connection  of  the  foot  of  a  knee  brace 
with  the  column  in  a  light  steel  building  frame.  The  force  F, 
24,000  Ibs.,  Fig.  in,  is  transferred  through  the  rivets  i,  2  and  3 
to  the  sketch  plate  and  in  turn  from  the  plate  by  the  rivets  4  to 
n,  inclusive,  to  the  column  angles.  The  center  of  gravity  of  the 
rivet  group,  4  to  n,  is  easily  located  as  its  center  of  symmetry.* 
Rivets  4,  7,  8  and  n  are  each  yj  ins.  from  the  center  of  gravity 
and  rivets  5,  6,  9  and  10  are  each  6.18  ins.  from  the  same  point. 


FIG.  in. 

The  rivets  must  offer  a  resistance  F^  equal,  opposite  and  par- 
allel to  FI.  This  resultant  force  Fz  acts  through  the  center  t)f 
gravity  of  the  rivet  group,  and  the  direct  force  F2  is  shared 
equally  by  the  rivets  of  the  group  so  that  each  rivet  carries 

— =  3000  Ibs.     The  forces  FI  and  F2  are  4.75  ins.  apart  so 

8 

that  the  moment  opposed  by  the  group  is  24,000  X  4.75   = 
114,000  in.  Ibs. 

The  force  opposed  by  a  rivet  will  be  proportional  to  its  dis- 
tance from  the  center  of  gravity  of  the  rivet  group.  The  rivets 

*  When  the  center  of  gravity  of  a  rivet  group  cannot  be  seen  by  inspection  it 
can  be  readily  calculated.  If  the  group  has  any  axis  of  symmetry  the  center  of 
gravity  must  lie  on  that  axis. 


TENSION,   COMPRESSION   PIECES   AND   BEAMS  87 

4,  7,  8  and  n  being  equally  distant  from  G  will  oppose  equal 
forces  and  similarly  of  the  rivets  5,  6,  9  and  10.  Let  S  be  the 

force  acting  at  4,  7,  8  and  n,  then      '—*  X  S   is  the  force  acting 

on  rivets  S,  6,  9  and  10.  The  moment  of  each  of  the  rivets 
4,  7,  8  and  n  will  be  5  X  7.50  and  the  moment  for  each  of  the 

rivets  5,  6,  9  and  10  will  be  (S  X  -^- )  X  6.18. 

V         7.507 

The  total  moment  for  the  eight  rivets  is 

(4  X  S  X  7.50)  +  (4  X  S  X  -^- )  =  114,000 in.  Ibs. 
V  7.507 

30  5  +  15.04  5  ==  114,000  in.  Ibs. 

0      114,000  .     „ 

S  =  -         -  =  2531  in.  Ibs. 

45-04 
The  force  acting  on  each  of  the  rivets  5,  6,  9  and  10  will  be 

(S  X  ^)  =  2531  X  —  =  2085  Ibs. 

V         7-50/  7-50 

The  forces  FI  and  F2  create  a  clockwise  moment  so  that  the 
resisting  moment  must  be  counterclockwise  and  the  forces  act 
to  produce  a  counterclockwise  moment  about  G.  The  forces 
act  at  right  angles  to  the  arms  or  lines  joining  the  rivet  centers 
with  G. 

The  resultant  shear  on  any  rivet  is  found  by  completing  the 
triangle  of  forces  at  each  rivet,  and  is  indicated  in  the  heavy  line. 
This  is  done  on  the  figure  for  the  r^a'..>JY 

several  rivets  and  the  maximum  shear 
comes  on  rivet  n  and  is  5530  Ibs. 
while  the  minimum  shear  acts  on 
rivet  4  and  is  850  Ibs.  1  6 

Should  there  be  no  axis  of  symme-  .  ~~f  "         1  £ 

try  take  any  two  base  lines,  preferably  *~  'Y 

at  right  angles,  and  find  the  distance 

of  the  center  of  gravity  from  each  of  these  base  lines.  Fig. 
112  represents  an  irregular  rivet  group.  The  first  base  line  is 


I 


88 


GRAPHICS  AND  STRUCTURAL  DESIGN 


taken,  passing  through  rivets  i  and  5.  The  rivets  are  assumed 
as  being  of  the  same  diameter  and  the  cross  section  of  each  rivet 
being  A ;  then  the  statical  moment  about  axis  x-x  will  be : 


Rivet. 

Area. 

Arm. 

Moment. 

I 

A 

O 

0 

2 

A 

a 

A  .a 

3 

A 

b 

A  .b 

4 

A 

c 

A  .c 

5 

A 

d 

A.o 

Total  area  =  5  A .         Total  Moment  =  A  (a  +  b  +  c) . 

The  distance  g  from  the  axis  x-x  to  the  center  of  gravity  will 
equal  the  statical  moment  divided  by  the  total  area  or 

=  A  (a  +  b  +  c) 

It  is  evident  from  this  expression  that  the  distance  g  equals  the 
sum  of  the  arms  a,  b  and  c,  etc.,  divided  by  the  number  of  rivets. 
In  the  same  way  the  distance  g'  from  the  axis  y-y  can  be  found 
and  the  center  of  gravity  is  then  located. 

In  designing  riveted  joints  the  following  well-established  rules 
are  used.  (These  do  not  apply  to  joints  in  boilers  or  cylinders 
where  maximum  efficiencies  are  desired.) 

i.   PREFERABLE  MINIMUM  DIMENSIONS,    INCHES 


Rivet  diameters. 

7 
8 

3 
4 

f 

i 

Center  to  center  of  rivets  *  

3 

a| 

2| 

i- 

Rivet  center  to  sheared  edge  

ll 

it 

It 

i 

Rivet  center  to  rolled  edge  

I 

7 
8 

*  Sometimes  taken  three  diameters  of  the  rivet. 

2.  The  maximum  pitch  in  the  line  of  the  stress  for  members 
composed  of  plates  and  shapes  should  be  6  ins.  for  f-in.  rivets; 
5  ins.  for  f-in.  rivets;  4^  ins.  for  f-in.  rivets;  and  4  ins.  for  J-in. 
rivets.  Where  angles  have  two  gauge  lines  and  the  rivets  are 
staggered  the  maximum  pitch  in  each  gauge  line  should  be  twice 
the  above  dimensions. 


TENSION,  COMPRESSION  PIECES  AND  BEAMS  89 

3.  Where  two  or  more  plates  are  used  in  contact  the  rivets 
holding  them  together  should  not  be  spaced  farther  apart  than 
12  ins.  in  either  direction. 

4.  The  pitch  of  rivets  in  the  direction  of  the  stress  should  not 
be  greater  than  6  ins.,  nor  exceed  16  times  the  thinnest  outside 
plate  connected,  and  not  more  than  50  times  that  thickness  at 
right  angles  to  the  stress. 

5.  The  maximum  distance  from  any  edge  should  be  8  times 
the  thickness  of  the  plate. 

6.  The  diameter  of  the  rivets  in  any  angle  carrying  calculated 
stresses  should  not  exceed  one-fourth  of  the  width  of  the  leg  in 
which  they  are  driven.     In  minor  parts  rivet  diameters  may  be 
|  in.  larger. 

7.  The  pitch  of  rivets  at  the  ends  of  built  compression  mem- 
bers shall  not  exceed  four  diameters  of  the  rivets  for  a  length 
equal  to  one  and  one-half  times  the  maximum  width  of  the 
member. 

8.  Two  pieces  riveted  together  should  always  be  secured  by 
at  least  two  rivets. 

9.  Joints  with  field-driven  rivets  should  have  from  25  to  50 
per  cent  more  rivets  than  would  have  been  required  for  shop- 
driven  rivets. 


CHAPTER  VII 
COLUMNS 

STRUCTURAL  engineers  generally  use  either  Rankine's  formula 
or  Johnson's  straight-line  formula  in  designing  columns.  The 
latter  is  a  modification  of  Rankine's  formula  and  the  results 
approximate  those  given  by  Rankine's  formula  within  the  usual 

working  limits  of       which  will  range  from  50  to   150.     The 

straight-line  formula  is  preferred  as  the  calculations  with  it  are 
simpler. 

The  following  give  these  formulae  in  their  usual  form  : 
Gordon's  or  Rankine's  formula  for  soft  steel, 


*+ 


, 

13,500  r2 
Gordon's  or  Rankine's  formula  for  medium  steel, 


1  1  ,000  r 
Johnson's  straight-line  formula  for  structural  steel, 

/=  16,000  -70--  (3) 

/  =  the  allowable  unit  compression  on  gross  section  of  col- 

umn in  pounds  per  square  inch. 
/  =  the  effective  length  of  the  column  in  inches.* 

*  The  effective  length  of  /  will  have  the  following  relations  to  L  the  total  length 
of  the  column: 

Both  ends  hinged  or  butting.  .  .  ...........  /  =  L 

Both  ends  fixed  .........................  l  =  \L 

One  end  fixed  and  one  end  hinged  .........  /  =  f  Z, 

One  end  fixed  and  other  end  free  ..........  I  =  2  L 

go 


COLUMNS  91 

r  =  the  least  radius  of  gyration  of  the  column  section  in  inches 


A 

I  =  the  least  moment  of  inertia  of  the  column  section  in  inches. 

A  =  the  area  of  the  column  section  in  square  inches. 

Ritter's  formula  most  nearly  meets  the  theoretical  require- 
ments but  it  is  not  much  used  as,  like  Rankine's  formula,  it  is 
cumbersome,  and  the  other  formulae  are  better  known.  It  has 
the  advantage  of  being  applicable  wherever  the  modulus  of 
elasticity  and  the  strength  of  the  material  at  the  elastic  limit  are 
known.  In  the  other  formulae  the  constant  for  the  material 
13,500  and  11,000  in  the  case  of  Rankine's  and  70  in  the  straight- 
line  formula  had  to  be  determined  experimentally.  Ritter's 
formula  as  generally  expressed  is: 


The  symbols  have  the  same  significance  as  just  given  for  formulae 
(i),  (2)  and  (3).     In  addition 

Sc  =  the  maximum  compressive  unit  stress  desired  on  the  con- 
cave side  of  the  column  in  pounds  per  square  inch. 
Se  =  the  unit  strength  of  the  material  at  its  elastic  limit  in 

pounds  per  square  inch. 
7T2  =  approximately  10. 

E  =  modulus  of  elasticity  of  the  material  in  pounds  per  square 
inch.     If  for  mild  steel,  the  following  values  may  be  assumed: 
E  =  30,000,000,     Se  =30,000    and    Sc  =  16,000, 

all,  pounds  per  square  inch.     With  these  values  substituted, 
formula  (4)  reduces  to 

, 16,000 16,000  ,  >, 

'  J  /1\  1  / 1\  9  \0/ 


10X30,000,000 

In  this  approximate  form  it  is  not  especially  cumbersome  but  it 
gives  values  slightly  under  those  calculated  by  formula  (3).    The 


92  GRAPHICS  AND  STRUCTURAL  DESIGN 

straight-line  formula  will  be  used  for  all  calculations  throughout 
the  text. 

The  American  Bridge  Company  have  issued  under  date  of 
Dec.  i,  1912,  new  specifications  for  structural  steel  work.  These 
specifications  recommend  for  columns  the  use  of  two  formulae, 

the  first,  /  =  19,000  —  100  -,  to  be  applied  to  values  of  -  up  to 
120;  the  second  formula  applies  only  to  secondary  members, 
those  permitted  to  have  -  values  ranging  from  1 20  to  200,  and  is 

/  =  13,000  —  50-  •     The  maximum  value  of  /  is  placed  at  13,000 

Ibs.  per  sq.  in. 

NOTE.  —  There  is  probably  no  section  of  Mechanics  of  Materials  about 
which  students  have  a  less  clear  conception  than  that  of  columns. 

It  is  here  assumed  that  the  student  has  this  fundamental  conception 
clearly  and  thoroughly  his  own.  Such  a  conception  is  essential  to  the 
intelligent  design  of  columns  and,  in  fact,  to  beams  also,  as  will  be  shown 
later.  Should  the  student's  idea  of  the  subject  be  in  any  way  vague  he  is 
urged  to  review  the  subject  in  any  Mechanics  of  Materials  with  which  he 
is  familiar. 

Problem.  —  Plot  the  values  of /given  by  formulas  (i),  (2),  (3) 
and  (5)  upon  cross-section  paper  for  values  of  -  between  50  and 

ISO- 
It  should  be  noted  in  the  several  formulae  that  r  is  the  least 

radius  of  gyration,  hence  that  section  will  theoretically  make  the 
best  column  where  the  radius  of  gyration  of  the  section  is  a  con- 
stant for  all  axes.  This  requires  a  circular  section,  which,  for 
economy  of  material,  would  generally  be  a  hollow  cylinder.  On 
this  account  rolled-steel  and  cast-iron  pipe  would  make  good 
columns,  but  that  cast  iron  is  being  but  infrequently  used  in 
important  structures  while,  owing  to  the  difficulty  of  making  con- 
nections to  them  by  brackets  or  framing,  round-steel  columns  are 
seldom  used. 


COLUMNS 


93 


One  of  the  simplest  columns  that  approaches  the  theoretically 
ideal  section  is  made  by  using  two  channels  spaced  so  as  to  make 
the  radii  of  gyration  of  the  column  section  referred  to  its  principal 
axes  approximately  equal.  The  columns,  Fig.  113,  are  then 
laced  with  flat  bars  riveted  to  the  channel  flanges  so  that  the  two 
channels  are  made  to  act  as  a  unit  throughout  the  column's 
length. 

/i  is  the  moment  of  inertia  of  one  channel  axis  i-i. 

1 2  is  the  moment  of  inertia  of  one  channel  axis  2-2. 

a   is  the  area  of  one  channel.    A  =  2  a. 


FIG.  113. 

The  inertia  of  the  total  column  section  about  axis  i-i  is 
L  -2/1. 

Since  r  =  y  -—  it  follows  that  if  the  radii  of  gyration  are  to  be 

equal  about  any  two  axes  of  a  section  the  inertias  must  be  equal; 
hence 

IC1  =  /s, 

from  which  

2 

It  should  not  be  understood  that  columns  must  always  ap- 
proach equal  radii  of  gyration  about  both  axes.  In  the  case  of 
columns  of  comparatively  short  lengths  it  may  be  cheaper,  owing 
to  the  reduction  in  the  work  of  manufacturing,  to  use  a  rolled 
section  than  a  built-up  column,  even  if  the  material  is  not  used 
to  such  good  advantage. 


94 


GRAPHICS  AND   STRUCTURAL  DESIGN 


It  therefore  frequently  happens  that  single  sections,  usually 
I  beams  or  angles,  make  good  columns.  In  some  cases  the  base 
is  merely  an  iron  casting  of  some  depth  for  stiffness,  containing 
a  pocket  cored  in  it  into  which  the  section  fits.  When  the  column 
is  brought  into  position  some  molten  soft  metal  is  poured  around 
it  to  hold  column  and  base  together.  In  more  important  cases, 
angle  and  plate  bases  similar  to  those  shown  in  Figs.  116  and 
117  are  used.  As  previously  stated  the  channel  column  is  one 


FIG.  114. 


FIG.  115. 


FIG.  116. 


FIG.  117. 


of  the  commonest  types.  These  are  illustrated  in  Fig.  113  and 
the  bases  for  such  a  column  are  shown  in  Figs.  116  and  117. 
The  flanges  of  the  channels  may  be  turned  in  instead  of  out, 
making  a  square  column,  but,  owing  to  the  difficulty  of  rivet- 
ing, it  costs  more  to  manufacture.  With  the  flanges  turned  out 
as  in  Fig.  113  the  lacing  may  be  replaced  by  plates.  Another 
design  uses  the  plate  on  the  outside  and  is  laced  on  the  inside, 
thus  permitting  of  inspection  and  painting.  The  sections  shown 
in  Figs.  114  and  115  are  also  accessible  for  painting.  In  Fig. 
115,  the  web  may  be  either  lacing  or  a  solid  plate.  This  section, 
when  deep,  makes  a  good  column  section  to  resist  combined 


COLUMNS  95 

compression  and  bending.     Its  strength  may  be  increased  by 
riveting  flange  plates  outside  the  angles. 

A  couple  of  the  simplest  forms  of  column  bases  are  shown  in 
Figs.  116  and  117.  The  proper  distribution  of  the  load  carried 
by  the  column  section  to  the  foundation  demands  not  only  an 
enlargement  of  the  foot  of  the  column  but  some  depth  to  assure 
its  proper  distribution. 

COMBINED  STRESSES 

In  a  column  the  force  may  act  parallel  to  the  axis  but  eccen- 
tric to  it.  Here  the  column  is  subjected  to  a  bending  moment 
P  (5  +  A)  where  P  is  the  force  acting  parallel  to  the  axis  and  A 
is  the  maximum  deflection  due  to  such  eccentric  loading.  The 
following  formula  by  Merriman  gives  the  maximum  resulting 
fiber  stress: 


A\_'   '  W*      1-5 

/r  =  maximum  combined  fiber  stress  in  pounds  per  square  inch. 
P  =  eccentric  force  on  column  in  pounds. 
A  =  area  of  the  column  section  in  square  inches. 
d  =  eccentricity  of  the  load  P  in  inches. 
e  =  distance  from  the  neutral  axis  to  the  extreme  fibers. 
r  =  radius  of  gyration  of  column  section  referred  to  the  axis 
about  which  the  bending  occurs. 
PP  PP 


I  =  length  of  the  column  in  inches. 
E  =  modulus  of  elasticity  in  pounds  per  square  inch. 

Where  a  piece  is  subjected  to  either  compression  or  tension, 
together  with  transverse  bending,  the  following  formula,  due  to 
Johnson,  is  commonly  used: 

Me 

' 


96  GRAPHICS  AND  STRUCTURAL  DESIGN 

fb  =  flexural  fiber  stress  in  pounds  per  square  inch. 
M  —  transverse  bending  moment  in  inch  pounds. 
e  =  the  distance  from  the  neutral  axis  to  the  extreme  fibers 

in  inches. 
7  =  moment  of  inertia  of  the  section  referred  to  the  axis  about 

which  the  bending  occurs. 
/  =  length  of  the  piece  in  inches. 
E  =  modulus  of  elasticity  in  pounds  per  square  inch. 
The  sign  (+)  is  used  when  P  puts  the  piece  in  tension,  the 
sign  (— )  when  compression  is  produced. 

To  this  flexural  fiber  stress  must  be  added  the  direct  com- 
pression (— )  or  the  tension  (+).  The  factor  given  as  10  varies 
with  the  character  of  the  loads  and  the  ends,  being  9.6  for  a 
simple  beam  uniformly  loaded,  but  1 2  for  a  similar  beam  with  a 
central  load.  Owing  to  its  simplicity  this  formula  is  frequently 
used  instead  of  the  first  formula. 

When  an  approximation  only  is  desired  the  stress  due  to  bend- 
ing may  be  found  by  solving  the  formula/  =  — - —  for  the  extreme 

fiber  stress  due  to  bending  and  adding  to  it  algebraically  the 
direct  stress  due  to  either  tension  or  compression.  This  method 
will  be  satisfactory  when  the  longitudinal  tension  or  compression 
is  not  large.  The  total  fiber  stress  should  not  exceed  that 
permitted  on  the  piece.  An  example  will  illustrate  the  use  of 
these  formulae. 

The  member  of  the  upper  chord  of  a  bridge  is  25  feet  long. 
It  weighs  no  Ibs.  per  foot,  and  has  the  following  section  (see 
Fig.  118): 

Top  plate  22  ins.  X  jV  in. 

Top  angles  3  ins.  X  3  ins.  X  f  in. 

Side  plates  16  ins.  X  f  in. 

Lower  angles  4  ins.  X  3  ins.  X  T7e  in. 

The  moment  of  inertia  axis  i-i  is  1003.8.     P  =  238,100  Ibs. 
The  distance  from  the  top  of  the  upper  plate  to  the  center  of 
gravity  of  the  built-up  section  is  6.30  ins. 


COLUMNS 


97 


The  bending  moment  due  to  the  dead  load  is 

WL      (no  X  25)  X  300  .     „ 

M  =  -  -**—  =  103,125  in.  Ibs. 

o  o 

The  fiber  stress  due  to  combined  bending  and  direct  stress  as 
given  by  the  approximate  method  is 

/  _  Me  _  103,125  X  6.30 

'  ~  ~T         1003.8 

Total  combined  stress 

.  238.100      0 
650  -h  -   7-^-  =  8190  Ibs. 


650  Ibs. 


63" 


FIG.  i i 8. 


FIG.  119. 


The  following  result  is  given  by  Johnson's  formula, 
/_       Me        _  103,125  X  6.3 


I  - 


PI2 


1003.8  - 


238,100  X  3QQ2 
10  X  30,000,000 


=  697  Ibs. 


Total  compression 


.   238.100      0        ., 
697  +       i6     =  8237  Ibs. 


The  following  example  will  illustrate  the  application  of  the 
formulae  when  the  load  is  parallel  to  the  longitudinal  axis  of  the 
piece  but  eccentric  to  it.  A  4  in.  X  4  in.  X  T76'm-  angle  84  ins. 
long  has  a  load  of  8800  Ibs.  applied  to  one  leg.  The  piece  is  in 
compression.  What  is  the  total  extreme  fiber  stress  in  com- 
pression? 

/  =  4.97;    area  =  3.31  sq.  ins.;  radius  of  gyration  =  1.23. 


98  GRAPHICS  AND   STRUCTURAL  DESIGN 

By  Merriman's  formula 

PI2  8800  X  i2o2 


48  El      48  X  30,000,000  X  4-97 


__  8800  ["     ,   1.16  X  1.16  i  +0.0177      1 

3.31  1_  i.232  i  -  (5  X  0.0177)]' 

fr  =  5300  Ibs. 

By  Johnson's  formula 
The  fiber  stress  in  bending  is 

Me 


/*- 


10  £ 


,         8800  X  1.16  X  1.16  „ 

/6=  -  —  -  -  -  —  =  2610  Ibs. 

8800  X   I202 

4.07  -- 

10  X  30,000,000 

The  fiber  stress  due  to  compression  is 

8800 
pc  —  -    ~  =  2660  Ibs. 

3-3i 

fr   =fb  +  fc  =   26lO  +  2660  =   5270  Ibs. 

In  this  case  the  maximum  stress  should  be  kept  under  that  allowed 
on  the  strut  as  given  by  a  column  formula;  hence 

/  =  16,000  —  (70  X  ^H  =  5360  Ibs. 
\          0.7 


LONG  BEAMS  UNSUPPORTED  LATERALLY 

When  a  beam  of  long  span  is  unsupported  laterally  the  upper 
flange  being  in  compression  is  liable  to  fail  as  a  column  by  buck- 
ling sideways.  There  is  no  very  satisfactory  theoretical  treat- 
ment of  this  subject.  The  compressive  fiber  stress  in  the  upper 
flange  is  usually  limited  by  formulae  or  rules  that  have  been  found 
safe.  Mr.  Christie,  basing  his  conclusions  upon  tests  made  on 
full-size  beams  for  the  Pencoyd  Iron  Works,  decided  it  was  safe 


COLUMNS  99 

to  use  a  desired  limiting  fiber  stress  up  to  a  span  of  twenty  times 
the  flange  width  and  that  from  this  point  the  working  fiber 
stress  should  be  uniformly  decreased  until,  at  a  span  of  70  times 
the  flange  width,  the  working  fiber  stress  should  be  one-half  the 
maximum  desired  stress. 

The  handbook  of  the  Cambria  Steel  Company  suggests  the 
following  formula, 

18,000 


Pi  = 


I  + 


3000  b2 
p=  allowable  compressive  fiber  stress  in  pounds  per  square 

inch. 

/=  length  of  span  in  inches. 
5=  width  of  flange  of  beam  in  inches. 

Mr.  C.  C.  Schneider  in  his  structural  specifications  limits  the 
allowable  compression  in  flanges  of  girders  to 

pi  =  16,000  —  200 -when  flange  is  composed  of  plates. 

pi  =  16,000  —  1 50  r  when  flange  is  a  channel. 
o 

In  this  case  the  fiber  stress  is  given  somewhat  lower  than  for 
beams,  which  is  consistent,  as  the  shallower  the  beam  the  more 
the  lower  flange,  which  is  in  tension,  reinforces  the  compression 
flange. 

The  1912  specification  of  the  American  Bridge  Company  limits 
the  span  unsupported  laterally  to  forty  times  the  flange  width, 
and  when  such  span  exceeds  ten  flange  widths  the  fiber  stress  is 

to  be  reduced  to  that  given  by  the  formula  19,000—3007- 

o 

The  following  curves,  Fig.  120,  are  based  on  formulae  derived 
by  the  author  in  an  attempt  to  explain  the  necessity  for 
reducing  the  fiber  stress  in  beams  where  the  beam  was  un- 
supported laterally.  The  derivation  of  these  formulae  is 


100 


GRAPHICS  AND  STRUCTURAL  DESIGN 


fully  explained  in  the  Proceedings  of  the  Engineers'  Club  of 
Philadelphia,  for  April,  1909. 


•no 


Hatio  of  Span  to  Mange  Wldjh. 

20  30  40 


Curves  give  the  <jc  of  the  desired  Maximum. 

Tiber  Stress  for  use  in  Designing  Long 

Beams  unsupported  Lateral  y. 


Q      !0      20      30      40       50      60      ?0      80      90     I'OO  ,J30    >20    130    140    150    .160    >70    180 
Ratio  oLSpan  to  Badius  of  Gyration  of. Compression  Flange 

FIG.  120. 


CHAPTER   VIII 
GIRDERS   FOR   CONVEYORS 

THE  half  plan  and  section  of  this  girder  are  given  in  Fig.  121, 
the  elevation  in  Fig.  122  and  the  end  view  in  Fig.  123. 
Assumed  Loading.  - 

Metal  at  105  Ibs.  per  lineal  foot 5*360  Ibs. 

Corrugated  steel  covering 2,000  Ibs. 

Foot-walk  and  sheathing 2,420  Ibs. 

Snow 6,120  Ibs. 

Total  dead  load 15,900  Ibs. 

Conveyor  load: 

Uniform  load  on  conveyor 3,900  Ibs. 

Weight  of  conveyor 3,700  Ibs. 

Total  moving  load  on  conveyor .  .     7,600  Ibs. 
Allowing  25  per  cent  additional  for  impact 1,900  Ibs. 

Total 9,500  Ibs. 

As  the  conveyor  loading  is  not  carried  equally  by  both  girders 
the  maximum  load  on  either  girder  must  be  calculated. 
The  total  maximum  load  on  one  girder  7950  +  6500  =  14,450. 

The  apex  load  is         ^    =  2060  Ibs. 
7 

Wind  Load.  —  The  horizontal  wind  load  is  assumed  at  20  Ibs. 
per  sq.  ft.  of  vertical  projection.  The  apex  wind  load  on  the 
upper  horizontal  girder  is  20  X  4  X  7^  ~  590  Ibs. 

The  stresses  must  now  be  determined  for  the  truss  under  the 
given  loading.  This  may  be  done  graphically  or  the  stresses 
may  be  found  very  readily  by  the  method  of  coefficients.  To 
illustrate  the  procedure  both  methods  will  be  used. 

The  method  of  coefficients  will  be  used  first.  For  the  explana- 
tion of  this  method  see  page  55.  These  calculations  are  made 


102 


GRAPHICS  AND   STRUCTURAL  DESIGN 


\ 


\ 


\ 


GIRDERS  FOR   CONVEYORS 


I03 


FIG.  124. 


Wtangfl 


Wtangfl 


FIG.  125, 


514 


DEAD  LOAD 
AND  LIVE  LOAD 

DIAGRAMS, 


J H 


FIG.  127. 


v. 


for  Fig.  125.    The  graphical  solution  for  the  main  truss  is  made 
in  Figs.  126  and  127. 

The  wind  bracing  and  graphical  analysis  are  given  in  Figs.  128 
and  129. 


104 


GRAPHICS  AND   STRUCTURAL   DESIGN 


0  =  45° 
Stress  EF 
Stress  EH 
Stress  EJ 
Stress  FD 
Stress  IB 
Stress  KA 
Stress  GH 
Stress  // 

1170* 


tang  45°  =  i  sec  45°  =  1.41 

=  —  3  X  2050  X  1.41  =  —  8500  Ibs. 
=  —  5  X  2050  X  i  =  —  10,250  Ibs. 
=  EK  =  -  6  X  2050  X  i  =  -  12,300  Ibs. 
=  +  3  X  2050  X  i  =  +  6150  Ibs.  =  stress  GC. 
=  +  5  X  2050  X  i  =  +  10,250  Ibs. 
=  +  6  X  2050  X  i  =  +  12,300  Ibs. 
=  +  2  X  2050  X  1.41  =  +  5420  Ibs. 
=  +  i  X  2050  X  1.41  =  +  2710  Ibs. 


590* 


FIG.  128. 


E-F-A 


FIG.  129. 
Selection  of  Members.  —  The  upper  chord. 

Maximum  compressive  stress,  dead  and  live  load 12,650  Ibs. 

Maximum  compressive  stress,  wind  load 3,660  Ibs. 

Total 16,310  Ibs. 

Try  2-3  in.  X  2  in.  X  i-in.  angles  spaced  J  in.  back  to  back, 
with  their  long  legs  parallel.     r2  =  0.88.     Their  length  is  88  ins. 

/        88 

-  =  -  -  =  loo. 

r      0.88 

The  allowable  fiber  stress  according  to  the  straight-line  formula 
is 

/i  =  16,000  —  70-  =  16,000  —  (70  X  100)  =  9000  Ibs. 


The  load  these  angles  will  carry  is  2  X  1.19  X  9000  =  21,420  Ibs. 


GIRDERS   FOR   CONVEYORS  105 

End  Strut  BF.  —  Its  length  is  120  ins.  and  its  load  8800  Ibs. 
Try  1-4  in.  X  4  in.  angle.     Its  radius  of  gyration  about  its  diag- 


onal axis  is  0.70.      -  =  --  =152.     The  allowable  fiber  stress 
r       0.79 

according  to  the  straight-line  formula  is 

/i  =  16,000  -  (70  X  152)  =  5360  Ibs. 

The  bending  moment  due  to  the  eccentric  loading,  when  the 
angle  is  assumed  as  T7g  in.  thick,  is  8800  X  i  :  i  =  10,032  in.  Ibs. 

The  fiber  stress  resulting  from  combined  bending  and  com- 
pression is  given  by  the  formula 


io  E 
see  page  95. 

10,200  X  1.16 
/  =  • —  =  2600  Ibs. 

8800  X   I202 

4.97 

io  X  30,000,000 

The  allowable  stress  according  to  the  straight-line  formula 
being  5360  Ibs.,  deducting  2600  Ibs.  leaves  2760  Ibs.  per  sq.  in. 
The  total  allowable  direct  stress  on  the  angle  then  is  3.31  X  2760 
=  91 50 Ibs.;  this,  exceeding  the  8800  Ibs.  acting  on  it,  the  angle 
is  satisfactory. 

Lower  Chord.  —  The  maximum  stress  is  16,310  Ibs.  Trying 
1-4  in.  X  3  in.  X  T\  in.,  and  placing  the  long  leg  horizontally,  its 
inertia  about  its  axis  parallel  to  the  short  leg  is  3.38;  the  dis- 
tance from  the  back  of  the  short  leg  to  the  axis  is  1.26  ins. 

The  bending  moment  due  to  this  eccentricity  is  16,310  X  1.26 
=  20,600  in.  Ibs. 

The  bending  moment  16,310  X  1.26  =  20,600  in.  Ibs. 

The  fiber  stress  due  to  bending  is 


10  £  10X30,000,000 


I06  GRAPHICS  AND   STRUCTURAL  DESIGN 

The  allowable  stress  on  the  net  section  is/  =  16,000  —  6850  = 
9150  Ibs. 

The  total  allowable  force  on  the  piece  then  is  (2.09  —  0.25)  X 
9150  =  16,850  Ibs. 

Vertical  HI.  —  This  is  a  compression  piece  carrying  2050  Ibs. 
1-3  in.  X  3  in.  X  J-in.  angle  will  be  tried. 

The  bending  due  to  eccentric  loading  is  M  =  2050  X  0.84  = 
1720  in.  Ibs. 

The  fiber  stress  due  to  this  bending  is 

.  My  1720  X  0.84 

/  =  T-PF  =  -  ~20.SoX88>      =I22olbs- 
7  ---  -      1.24  -- 

10  £  10  X  30,000,000 

The  allowable  compression,  according  to  the  straight-line 
formula,  is 

7  ,  OQ     \  - 

/  =  16,000  —  70-  =  16,000  —  (  70  X  -  -)  =  5600  Ibs. 
r  \          0.597 

The  net  compression  allowed  per  square  inch  of  the  section 
is  5600  —  1220  =  4380  Ibs. 

The  area  of  the  angle  multiplied  by  4380  is  1.44  X  4380  = 
6300  Ibs.  Hence  the  3  in.  X  3  in.  X  J-in.  angle  is  more  than 
ample. 

Diagonals.  —  The  maximum  tension  in  the  diagonals  is  6000 
Ibs.  Try  i-2j  in.  X  2  in.  X  J-in.  angle.  Its  inertia  about  an 
axis  through  its  center  of  gravity  and  parallel  to  its  long  leg  is 
0.37.  The  distance  from  the  back  of  the  angle  to  the  above 
axis  is  0.51  in. 

The  bending  moment  due  to  the  eccentric  application  of  the 
load  to  the  angle  is  6000  X  0.51  =  3060  in.  Ibs. 

The  extreme  fiber  stress  due  to  this  bending  is 
-          My  3060  X  0.51 


0.37 


,    ,, 

6000       =  236olbs- 


. 
10  £  10X30,000,000 

The  allowable  stress  per  square  inch  of  the  net  section  is  16,000 
—  2360  =  13,640  Ibs.     Net  section  X  13,648  =  (1.07  —  0.20)  X 


GIRDERS   FOR   CONVEYORS  107 

13,640   =   11,850  Ibs.     Tnis  is  satisfactory,  being  the  smallest 
angle  we  can  use. 

Wind  Bracing.  —  The  maximum  stress  in  the  diagonals  is 
2550  Ibs.  The  smallest  flat  allowed  being  2  X  i  in.,  and  as  this 
will  carry  [(2  X  0.25)  —  0.20]  X  16,000  =  4800  Ibs.  (the  net  area 
times  the  allowable  fiber  stress),  it  is  ample. 

STRUT  IN  UPPER  HORIZONTAL  BRACING 


57 

The  bending  moment  under  the  load  between  RI  and  R2  is 
M  =  400  X  12  =  4800  Ibs. 


FIG.  130.  FIG.  131. 

Trying  3  in.  X  3  in.  X  J-in.  angles,  their  inertia  about  an  axis 
through  their  center  of  gravity  and  parallel  to  a  leg  is  1.24,  the 
distance  from  the  back  of  the  angle  to  this  axis  is  0.84  in.,  the 
radius  of  gyration  referred  to  this  axis  is  0.93,  while  the  radius 
of  gyration  referred  to  the  diagonal  axis  is  0.59. 

I  QQ 

-  for  the  full  length  =  -    -  =  97. 
r  0.93 

-  for  the  length  of  57  ins.   =  -^—  =  97. 
r  0.59 

The  allowable  stress  /  =  16,000  —  f  70  X  -J  =  16,000  —  (70  X  97) 
=  9210  Ibs. 

The  fiber  stresses  due  to  the  bending  of  4800  in.  Ibs.  in  57-in. 
span  are, 

,      Me     4800  X  0.84                        ,      4800  X  2.16      0 
/«  =  -f~ ~  =  325° lbs-    fc  =  -   =  8350  Ibs. 

/  1.24  I.2A 


108  GRAPHICS  AND   STRUCTURAL  DESIGN 

The  bending  due  to  the  eccentric  loading  as  a  column  will  be 
My  (i  180  X  0.84)  X  0.84         835  _6Solb5 

~_*L      It24  "8QX578  1-23 

10  X  30,000,000 


.       680  X  2.16 
ft  =  -  o  -  =  1750  Ibs. 
0.04 

The  direct  compression  =  force  divided  by  the  area  of  the 

1180      0      „ 
section  =  --  =  820  IDS. 

1.44 

The  maximum  compression  will  be  9170  Ibs.,  which,  being 
under  the  9210  Ibs.  allowed  the  section,  is  satisfactory. 

The  beams  serving  as  struts  in  the  lower  horizontal  bracing 
carry  a  total  load  of  1080  Ibs., 


-f  400  (snow  load)  =  1080  Ibs. 

R>=     I08°  X  57  =  685  lbs. 
90 

M  =  685  X  33  =  22,605  in.  lbs. 

Assuming  the  channel  unsupported  laterally  for  its  length  of 
88  ins., 

88 

length  -T-  flange  width  =  -  —  =  45- 

1.92 

Allowable  fiber  stress  about  11,000  lbs.  per  sq.  in., 

M      I      22,605 

—  =  -  =  —i—  ±  =  2.06. 

/       e      ii  ,000 

This  suggests  about  a  5  -in.  channel. 

The  direct  compression  being  1  180  lbs.  and  applied  at  the  upper 
flange  the  moment  due  to  eccentric  loading  is  If  =  1180  X  2.5  = 
2950  lbs. 


10X30,000,000 


GIRDERS   FOR   CONVEYORS  109 

Direct  stress  —    -  =  605  Ibs. 

i-95 
The  extreme  compressive  stress  due  to  bending  of  a  5-in.  U  is, 

.      Me      22,605 
/  =  —      — ^    =  7535  Ibs.  per  sq.  in. 

Total  compressive  stress  7535  +  570  +  605  =  8710  Ibs.,  being 
below  11,000,  is  satisfactory. 


CHAPTER    IX 
TRUSSES,   BENTS   AND    TOWERS 

IT  is  frequently  necessary  around  works  to  carry  large  gas  or 
air  mains  overhead.  These  mains  are  sometimes  lined  with  brick 
to  prevent  radiation  loss  from  the  hot  gas  or  air.  There  is 
consequently  considerable  load  per  foot,  and  as  long  spans  are 
preferable,  thus  avoiding  too  large  a  number  of  posts  or  bents 


FIG.  132. 

that  would  block  up  the  yard,  the  pipe  must  be  trussed.  Fig. 
132  represents  such  a  trussed  pipe.  Care  should  be  taken  that 
any  load  transferred  to  or  from  the  pipe  should  be  distributed 
over  the  pipe  by  a  saddle  and  if  necessary  the  pipe  should  be 
reinforced  at  such  points.  In  Fig.  132,  AB  is  the  reaction  in 
the  post,  AC  the  compression  in  the  pipe,  and  BC  the  tension 
in  the  rods. 

The  span  over  which  the  pipe  section  acts  as  a  beam  is  -. 

O 

Steam,  air,  gas  and  water  pipes  may  be  carried  upon  a  light 


TRUSSES,  BENTS   AND   TOWERS 


III 


trussed  bridge  somewhat  resembling  the  conveyor  truss  (see 
page  102),  or  may  be  hung  from  a  cable,  as  shown  in  Fig.  133. 

The  stress  in  the  wire  rope  depends  upon  the  load  carried  and 
upon  the  sag  permitted  in  it.  As  the  weight  of  the  rope  will  be 
small  compared  with  the  load,  the  effect  of  the  weight  of  the 
rope  may  be  neglected.  In  Fig.  133,  having  assumed  the  rope 
diameter,  lay  off  R  representing  the  maximum  desired  pull  on 
the  rope,  then  draw  the  horizontal  and  vertical  components  H 
and  V.  Taking  the  forces  acting  about  apex  i,  the  horizontal 

FIG.  133. 


FIG.  134. 

component  of  BC  must  equal  the  horizontal  component  of  CA. 
Similarly,   the  horizontal  components  of  the  forces  acting  at 

W  W 

apex  2  are  equal.    Now  in  Fig.  134,  XD  =  --  and  DA   =  —  • 

4  ° 

H  is  the  horizontal  component  of  the  force  BC  acting  in  the 
guy  rope.  CA  and  CD  give  the  magnitudes  and  directions  of 
the  rope  stresses  and  the  sag  of  the  rope  may  be  found  by  drawing 
the  strings  in  Fig.  133,  corresponding  to  these  rays  in  Fig.  134. 
Should  the  sag  prove  objectionable  a  heavier  rope  with 
greater  stress  or  a  greater  stress  may  be  used  in  the  first  rope  if 
permissible. 

W 
The  vertical  load  on  the  post,  Fig.  133,  is  V  -\ 


112 


GRAPHICS  AND   STRUCTURAL  DESIGN 


Bents,  Fig.  135,  may  be  used  to  support  both  pipes  and  wires. 
If  the  bent  is  an  end  one  and  guyed  the  vertical  load  will  be 
determined  as  in  the  preceding  example.  If  it  is  an  intermediate 
bent  it  will  not  carry  the  vertical  component  of  the  guy  rope  but 
merely  its  proportion  of  the  weight  of  the  wires  and  pipes.  It 
may  also  be  subjected  to  wind  load,  in  which  event  the  stresses 
in  the  bracing  can  be  determined  and  these  pieces  selected  for 
their  respective  loads,  see  Fig.  137.  The  wind  acting  on  cylin- 
drical surfaces  is  commonly  assumed  as  -fa  of  the  pressure  that 
would  act  on  the  vertical  projection  of  that  surface. 


FIG.  135. 


DEAD 

LOAD 

STRESS 

DIAGRAM 


FIG.  138. 


As  the  loads  are  generally  very  light  and  not  subjected  to 
shock,  the  values  of  -  are  permitted  to  reach  180  or  200. 

As  the  wind  load  is  transmitted  to  the  truss  proper  by  pieces 
subjected  to  bending,  the  diagram,  Fig.  136,  has  been  started  by 
connecting  the  wind  force  to  the  truss  through  the  dotted  or 
substituted  members.  The  dead-load  stress  diagram  is  shown 
in  Fig.  138. 

Towers  for  Transmission  Lines.  -  -  There  is  no  standard 
practice  but  the  four-angle  type  is  the  most  common.  The 


TRUSSES,    BENTS   AND   TOWERS  113 

loading  upon  these  poles  consists  of  dead  load  of  pole  and  weight 
of  wire,  also  ice,  which  may  add  materially  to  this  direct  load 
in  localities  visited  by  sleet.  A  live  load  due  to  the  wind  acts 
at  right  angles  to  the  dead  load,  and  in  addition  there  may  be  a 
direct  pull  along  the  line  caused  by  a  wire  or  wires  breaking. 
The  ice  load  is  generally  assumed  as  a  |-in.  coating  around  the 
wire. 

Wind  loads  assumed  as  acting  on  wires  vary  considerably. 
Mr.  R.  Fleming  in  an  article  in  Engineering  News,  Nov.  28,  1912, 
recommends  30  Ibs.  per  sq.  ft.  of  exposed  surface  upon  the  tower, 
15  Ibs.  per  sq.  ft.  of  projected  area  of  the  bare  wire,  reduced  to 
10  Ibs.  for  ice-coated  wires.  He  further  recommends  that  poles 
carrying  three  wires  be  designed  to  resist  the  unbalanced  pull,  due 
to  one  wire  breaking,  while  for  six  wires  two  of  them  shall  be 
considered  as  breaking.  The  unbalanced  load  is  the  force  that 
breaks  the  wire  and  the  following  percentages  of  the  ultimate 
strengths  of  the  wires  are  recommended :  the  full  ultimate  strength 
up  to  and  including  a  No.  4  wire;  a  No.  3  wire,  90  per  cent;  a 
No.  2  wire,  80  per  cent;  a  No.  i  wire,  70  per  cent;  a  No.  o  wire, 
60  per  cent;  and  for  all  larger  wires,  50  per  cent.  The  ultimate 
strength  per  square  inch  of  hard-drawn  copper  wire  is  from 
50.000  to  65,000  Ibs.  The  towers  are  designed  to  meet  the 
requirements  of  the  worst  combination  of  these  conditions. 

Using  ordinary  structural  steel,  Mr.  Fleming  recommends  the 
following  formulae  for  columns: 

/c  = *~z —  ,  for  the  most  exacting  city  service. 

I+  18,000  r2 

r  27,000,  r        , 

fc  = - — ,  for  less  exacting  service. 

18,000  r2 

In  estimating  the  loading  upon  the  poles  the  additional  load- 
ing due  to  the  line  running  on  an  incline  or  rounding  a  curve 
should  not  be  overlooked.  This  is  generally  provided  for  by 


114  GRAPHICS  AND   STRUCTURAL  DESIGN 

placing  the  poles  closer  together  in  these  places  rather  than 
changing  the  tower  design. 

In  estimating  the  tension  in  the  line  and  length  of  the  wire 
between  supports  it  is  common  practice  to  assume  the  curve  of 
the  line  a  parabola  which  differs  but  slightly  from  the  catenary 
and  is  much  more  readily  computed. 

S  =  distance  between  poles  in  feet. 

L  =  length  of  wire  in  feet,  measured  along  the  curve. 

d  —  sag  or  drop  in  wire  in  feet,  measured  midway  between 
poles. 

H  =  tension  in  lowest  point  of  curve  in  pounds. 
c  =  tension  constant. 

w  =  resultant  load,  measured  in  pounds  per  foot  of  wire.  In 
winter  the  weight  of  wire  and  ice  acting  vertically 
and  the  wind  acting  horizontally.  In  summer  the 
weight  of  wire  acting  vertically  and  the  wind  on  the 
bare  wire  acting  horizontally. 


(4) 

For  a  more  complete  mathematical  discussion  of  this  subject, 
see  Bulletin  No.  54  of  the  University  of  Illinois.  "Mechani- 
cal Stresses  in  Transmission  Lines,"  by  A.  Guell.  Jan.  22, 
1912. 

The  length  of  the  wire  at  the  summer  temperature  will  increase 
due  to  the  temperature  but  will  have  less  extension  due  to  the 
loading  as  the  vertical  load  will  not  include  the  weight  of  the  ice 
and  the  wind  will  not  act  on  so  great  a  surface.  Mr.  Guell  gives 
the  following  formulae: 


— 


TRUSSES,   BENTS   AND   TOWERS 


SECTION  A-A 


'Concrete.! 


1%  Bolt 

Top  upset  to  %%' 


FIG.  139. 


Il6  GRAPHICS   AND   STRUCTURAL  DESIGN 

Here 

LI  =  length  of  the  wire  along  the  curve,  measured  in  feet, 

at  the  summer  temperature. 

L  =  length  of  the  wire  for  winter  temperature  and  loading. 
a  =  coefficient  of  expansion  per  degree  Fahr.  =  0.00000956. 
/  =  temperature  range,  degrees  Fahr. 
H  =  tension  at  low  point  in  curve  in  winter,  pounds. 
HI  =  tension  at  low  point  in  curve  in  summer,  pounds. 
A  =  area  of  wire,  square  inches. 

E  =  modulus  of  elasticity,  pounds  per  square  inch.  For 
hard-drawn  copper  wire,  12,000,000  to  16,000,000  Ibs. 
per  sq.  in. 

The  stresses  in  the  wire  will  be  lower  in  summer  than  in  winter, 
but  it  is  necessary  to  estimate  the  sag  for  the  summer.  It  may 
be  approximated  as  follows.  In  equation  (5)  estimate  the  value 
of  Li,  omitting  the  last  term,  as  HI  is  not  known.  A  preliminary 
estimate  of  HI  may  be  made  by  assuming 

HI  =  Wi 

H       w' 
Here 

Wi  =  resultant  load  per  foot  in  summer. 
w   =  resultant  load  per  foot  in  winter. 

This  value  of  HI  would  then  be  substituted  in  equation  (5)  and 
the  last  term  need  not  be  discarded.  Having  now  found  LI 
substitute  this  value  for  L  in  equation  (i)  and  solve  for  c.  This 
value  of  c  used  in  equation  (2)  gives  an  approximate  value  of  d. 
Now  in  equation  (4),  w  is  known  from  the  loading  and  H  can  be 
found.  This  is  an  approximation  to  HI  required  in  equation  (5) 
and  the  value  of  LI  may  now  be  revised  by  placing  this  value 
of  HI  in  this  equation.  The  trials  may  be  repeated  until  the 
desired  degree  of  approximation  is  reached. 

Cross  arms  should  be  designed  for  a  minimum  vertical  load 
of  from  looo  to  1200  Ibs.,  also  for  an  unbalanced  horizontal 
pull  due  to  the  wires  breaking  and  estimated  as  previously  given. 


TRUSSES,  BENTS   AND  TOWERS  117 

The  twisting  of  the  tower  due  to  such  unbalanced  loading  should 
not  be  overlooked. 


! 

4- 


Concrete 


FIG.  140. 

Figure  139  shows  one  type  of  tower  and  foundation  and  follows 
a  design  by  Westinghouse,  Church,  Kerr  &  Co.,  as  does  also 
Fig.  140,  which  illustrates  a  type  of  foundation  with  greater 
base  area. 


CHAPTER  X 
DESIGN   OF   A    STEEL   MILL  BUILDING 

THE  design  of  a  steel  mill  building  as  generally  carried  out 
requires  considerable  experience  and  judgment  on  the  part  of 
the  designer.  The  truss  stresses  are  usually  determined  for  an 
equivalent  load  rather  than  by  the  separate  determination  of  the 
dead-,  snow-  and  wind-load  stresses  and  their  combination  for 
maximum  effect.  The  following  equivalent  loads  may  be  used 
on  spans  under  80  ft. 


Kind  of  roof. 

Lbs.  per  sq.  ft. 

Gravel  or  composition  roofing 
Gravel  or  composition  roofing 
Corrugated  steel  sheeting 

on  boards 

45-50 
60 
40 
50 
65 

on  3~in.  concrete. 

Slate  on  boards 

Slate  on  3-in.  concrete,  flat..  .  . 

Where  no  snow  need  be  considered  these  figures  can  be  reduced 
by  10  Ibs.,  excepting  that  no  roof  shall  be  designed  for  less  than 
40  Ibs.  per  sq.  ft. 

Stiffening  the  Structure.  —  The  structure  is  stiffened  first 
where  possible  by  introducing  a  knee  brace,  running  from  a 
panel  point  in  the  lower  chord  of  the  truss  to  a  point  as  low  as 
convenient  on  the  column.  Where  the  column  carries  a  crane 
runway  this  knee  brace  must  generally  be  omitted  and  is  usually 
replaced  by  a  knuckle  brace  which  stiffens  the  connection  be- 
tween the  column  and  the  truss. 

The  structure  is  also  braced  in  the  plane  of  the  lower  chords  of 
the  truss  to  hold  the  tops  of  the  columns  at  constant  distances 
apart,  and  in  some  cases  to  carry  the  wind  forces  acting  along 
the  side  of  the  building  to  the  transverse  bracing  at  its  ends. 

118 


DESIGN  OF  A  STEEL  MILL  BUILDING  119 

To  prevent  the  trusses  rotating  about  their  lower  chords, 
bracing  is  placed  in  the  plane  of  the  upper  chords.  This  bracing 
is  commonly  omitted,  at  least  in  alternate  bays. 

To  resist  the  wind  pressure  on  the  ends  of  the  buildings  and 
the  cumulative  pressure  along  the  roof,  together  with  any  crane 
thrust,  lateral  bracing  is  placed  in  the  plane  of  the  building 
columns.  This  bracing  usually  consists  of  an  eave  strut  which 
is  frequently  a  light  latticed  girder  composed  of  four  angles,  the 
web  being  composed  of  diagonal  lattice  bars. 

Where  possible  the  diagonal  bracing  should  extend  to  the  ends 
of  the  columns.  Where  the  column  height  is  considerable  an 
additional  strut  is  placed  about  half  way  up  the  column.  Both 
eave  strut  and  intermediate  strut  run  the  entire  length  of  the 
building  and  fasten  to  each  column.  The  diagonal  braces  are 
commonly  placed  at  the  end  bays  and  then  at  intervals  along 
the  building  as  deemed  necessary  by  the  designer.  Where  the 
bracing  cannot  be  extended  to  the  column  bases  the  columns 
must  be  designed  to  resist  the  bending  due  to  the  horizontal 
forces  resulting  from  the  wind  acting  on  the  end  of  the  building 
and  any  crane  thrust.  This  bending  will  be  influenced  by  the 
way  the  column  bases  are  secured  and  may  be  treated  similarly 
to  the  transverse  bent,  page  45. 

The  transverse  bracing  in  the  ends  of  the  building  is  intended 
to  resist  whatever  of  the  forces  acting  on  the  side  of  the  building 
from  wind  pressure,  or  thrust  on  the  columns  from  the  crane, 
may  be  transferred  to  the  ends  by  the  structure. 

In  the  simpler  buildings  it  is  possible  to  assume  certain  exter- 
nal forces  resisted  by  a  particular  bracing  truss  and  then  make 
the  design  accordingly.  In  the  more  complicated  buildings  the 
designer  modifies  his  calculations  by  what  has  been  previously 
found  to  be  satisfactory. 

The  following  sketches  illustrate  one  general  plan  of  steel  mill 
bracing,  Figs.  141  to  143.  See  also  Figs.  156  to  158. 

A  number  of  types  of  roof  trusses  are  shown  in  Figs.  144 
to  151. 


120 


GRAPHICS   AND   STRUCTURAL   DESIGN 


TRANSVERSE  END 

BENT  BRACING 


FIG.  141. 
ELEVATION! 


^"^^'' 

Purlin 

^_ 

^-' 

**^~^^''' 

^^-'     ^-^ 

" 

^-•' 

^^^ 

^'"    "~^^ 

^•^    „-•' 

^^J'*-'" 

"^•v^    ^x-' 

1      Purlin 

^^^ 

^''    ^^ 

^^x''"^^- 

^^"        ^v.^ 

.. 

^ 

^    „'" 

"^^j-  ^' 

*^^         ^^" 

^ 

*"'  "^ 

Eave 

^.--"   ^-^ 

,^'""^-^ 

•• 

~~-^ 

X 

Strut 
Strut 

V 

/            \ 

^X                   &1I 

^x           / 

E 

, 

/\^ 

: 

/ 

A- 

X 

\                       X 

X 

T 

^1^'' 

- 

N 

NNXX/  ~ 

1 

/       ^^ 

/           \ 

FIG.  142. 

Eave  Strut 


Lower 
Chord 
Bracing 


\ 


Purlin 


3rd  Brae  ug 


Purlin 


Eave  Strut 

PLAN 
FIG.  143. 


DESIGN  OF   A   STEEL   MILL   BUILDING 


121 


The  number  of  panels  into  which  the  upper  chord  should  be 
divided  will  depend  upon  the  span  of  the  truss  and  upon  the  roof 
covering  or  sheathing.  In  the  case  of  corrugated-steel  roofing, 
it  is  desirable  that  the  sheets  should  extend  over  three  purlins. 
That  the  roof  covering  be  amply  stiff  it  is  necessary  with  the 
ordinary  gauges  to  limit  the  distance  between  purlins  to  from 


FIG.  144. 


FIG.  145. 


FIG.  146. 


FIG.  147. 


FIG.  148. 


FIG.  149. 


FIG.  150. 


FIG.  151. 


4  to  6  ft.;  consequently,  the  purlins  are  generally  spaced  about 
4  ft.  In  the  case  of  other  roof  coverings  stiffness  is  also  re- 
quired, but  as  most  of  these  coverings  are  laid  upon  sheathing 
the  purlins  can  be  spaced  farther  apart  providing  the  sheathing 
is  made  correspondingly  thicker.  If  the  upper  chord  is  not  in- 
tended to  take  bending  it  is  necessary  to  have  a  member  of  the 
truss  connected  to  the  upper  chord  under  each  purlin. 


122 


GRAPHICS  AND   STRUCTURAL  DESIGN 


Frequently,  however,  the  upper  chord  is  designed  of  a  plate  and 
two  angles  forming  a  tee-section.  In  this  case  the  section  is 
calculated  to  resist  bending  and  the  purlins  can,  therefore,  be 
spaced  as  desired,  and  a  less  complicated  truss  used. 

The  pitch  of  the  roof  will  vary  with  the  character  of  the  roof 
covering.     The  following  table  gives  the  usual  pitches. 


Kind  of  roofing. 

Minimum  pitch. 

Usual  pitch. 

Corrugated,  steel 

i 

1 

Slate  or  tile 

i 

M 

Shingles                              .           .... 

1 

V 

Gravel                                

Maximum  pitch 

fin  12" 

Asphalt                             

i 

fin  12" 

Patent  compositions  

1 

i 

Design  of  a  Steel  Mill  Building,  Figs.  152  to   158. —  The 
truss  will  be  designed  for  an  equivalent  load  of  40  Ibs.  per  sq.  ft. 


FIG.  159. 

of  horizontal  projection  of  the  roof.  The  rise  will  be  made  one 
in  four,  and  the  trusses  will  be  spaced  17  ft.,  the  span  being 
86  ft.  o  ins.  Total  load  per  truss  =  86  X  17  X  40  =  58,480  Ibs. 

Tan-1  \  =  26°  30'.     In  Fig.  159  the  length  of  rafter 


=  47.1  ft.     Distance  between  apices 


Sill  2O      3 

=  9.42  ft.    In  this  way 


the  lengths  of  the  several  members  are  found  to  be 


DESIGN  OF  A   STEEL  MILL  BUILDING  123 

CH,  DI,  EK,  FN  and  GO  =  9.42  ft. 

m  =  ^  =  4.71  ft. 

2 

JK  =  2  X  4-71  =  942  ft. 


LM  =  3  X          ==  14-13  ft. 


NO  =  =  7.07  ft. 

2 

DESIGN  OF  MEMBERS.    COMPRESSION  PIECES 
Member  HI.  —  The  stress  in  HI  is  5200  Ibs.  and  is  given 
by    Fig.    153.     The    value    of  -  =  120   or   r  =      -  =  4.71  X 

—  =  0.47.     Trying  a  3  in.  X  i\  in.  X  i-in.  angle,  which  is  the 

I2O 

smallest  angle  allowed,  its  net  area  is  1.32  sq.  ins.  less  0.20 
sq.  in.  for  one  rivet  hole  or  1.12  sq.  ins. 

-  =  4.71  X  -   -  =  106. 
r  0.53 

The  allowable  unit  stress  on  this  column  is 

/=  16,000  —  70  -  =  16,000  —  (70  X  106)  =  8580  Ibs. 

The  total  force  allowed  on  the  piece  is  1.12  X  8580  =  9610  Ibs. 
This,  being  well  above  the  stress  in  the  piece,  is  satisfactory. 
Member  JK.  —  The  length  of  JK  is  9.42  ft.;  the  stress  in  it 

/  12 

is  7800  Ibs.     Limiting-  to  120  as  before,  r  =  9.42  X  -  -  =  0.94. 
r  1  20 

Trying  two  3  in.  X  i\  in.  X  i-in.  angles,  their  net  area,  after 
deducting  for  rivet  holes  in  one  leg  of  each  angle,  is  (2.64  — 
0.4)  =  2.24  sq.  ins.  The  value  of  r  with  the  long  legs  back  to 
back  is  0.95.  For  this  value  of  r  we  find  the  allowable  unit 
stress  to  be 

/  =  16,000  —  70-  =  16,000  —  (70  X  120)  =  7600  Ibs. 


124  GRAPHICS  AND   STRUCTURAL   DESIGN 

The  total  load  these  angles  will  carry  then  is  2.24  X  7600  = 
17,000  Ibs. 

Member  LM.  —  The  length  of  this  member  is  14.13  ft.     Its 
load  is  13,000  Ibs.     The  least  radius  of  gyration  is 

14.13  X  12 

r  =  -  -  =  1.41. 

120 

The  load  being  very  light  two  4  in.  X  3  in.  X  J-in.  angles  will  be 
tried,  although  the  radius  of  gyration  is  low. 

=      6 


/  =  16,000  —  -  —  =  16,000  —  9520  =  6480  Ibs. 


r  1.25 

so  that 

—  -  —  = 

The  load  on  two  angles,  allowing  one  rivet  hole  in  each  angle,  is 
load  =  2  (1.69  —  0.20)  X  6480  =  19,450  Ibs. 

Member  NO  is  slightly  shorter  than  JK  and  carries  a  lighter 
load,  but  will  be  made  of  the  same  angles,  3  in.  X  2^  in.  X  i  in. 

Members  MA,  JA,  LA  and  PA  are  in  tension  and  will  be 
made  alike. 

The  maximum  stress  in  these  members  is  52,400  Ibs.  Trying 
two  3  in.  X  3  in.  X  f-in.  angles  their  net  section  is  (4.22  —  0.60) 
X  16,000  =  58,000  Ibs.  This  is  sufficiently  close  to  the  section 
required. 

Member  OP.  —  This  is  a  tension  piece  carrying  17,000  Ibs. 
stress.  Trying  two  3  in.  X  2^  in.  X  J-in.  angles  and  allowing  for 
one  rivet  hole  they  will  carry  2  (1.32  —  0.20)  X  16,000  =  35,800 
Ibs.  This  is  more  than  sufficient  but  two  angles  are  preferable 
here  and  these  angles  are  the  smallest  ones  allowed  by  the 
specifications. 

The  less  important  tension  members  may  be  made  of  a  single 
angle.  A  3  in.  X  2^  in.  X  i-in.  angle  is  the  smallest  permitted 
and  it  will  carry  (1.32  —  0.20)  X  16,000  =  17,900  Ibs.  This 
angle  will  be  used  for  the  members  //,  KL  and  MN. 


DESIGN  OF  A  STEEL  MILL  BUILDING  I2«; 

Upper  Chord.  —  The  upper  chord  will  be  made  of  the  section 
shown  in  Fig.  160.  It  is  first  necessary  to  find  the  center  of 
gravity,  and  then  the  moment  of  inertia  of 
this  section. 

Area,  sq.  ins.  Moment. 

2   X  2.09     =  4.18  X  0.76   =      3.18 
9  X  0.313  =  2.82  X  4-50  =     2.65 


x  = 


7  .  oo  5  .  83 

=  2.  26  ins. 


7.00 


The  inertia  of  the  combined  section  then 
follows: 

Angles,  2  X  1.65  =     3.30 

Angles,  Ah2,  2  X  2.09  X  (2.26  —  0.76)2  =     9.42 

Plate,  g- 


FIG.  160. 


=,9.oo 


Plate,  Ah2  =  2.81  X  (4.50  -  2.26)2 


=    14.08 
7  =    45.80 

The  span  of  CH  is  9.42  ft.     When  two  purlins  are  used  in  each 

5850 


span  of  9.42  ft.  the  purlin  load  will  be 


2925  Ibs.     Assum- 


ing the  purlins  centrally  located  on  the  above  span  the  bending 
moment  is 


=     X  2925  X 


o  \  4   '      o  4 

The  extreme  fiber  stress  in  compression  is 


5I)6ooin.  Ibs. 


Me 


X 


2.26 

7~ 
45.0 


while  the  extreme  fiber  stress  in  tension  is 


ft  =  —T  =  5J>6oo  X 

/  45.00 

Combining  these  fiber  stresses  we  have, 
Total  fiber  stress  in  compression  side, 

-  2540  - 


254olbs., 


7600  Ibs. 


=    -  10,940  Ibs. 


Total  fiber  stress  in  tension  side,  7600  —  P  '       j  =  —  800  Ibs. 

\  6.99  / 


126  GRAPHICS  AND   STRUCTURAL  DESIGN 

The  radius  of  gyration  of  this  section  must  now  be  found 
about  the  axis  through  the  plate  and  parallel  to  the  short  legs 
of  the  angles. 

Inertia  of  two  angles  referred  to  axis  2-2,  2  X  3.38  =    6.76 
Ah2  of  the  angles,  2  X  2.09  X  (1.26  +  0.155)2       =    8.36 

_         /  =  15-12 

l~f         l  — 
The  radius  of  gyration  =  r  =  y  —  =  y-     -=1.44, 

*b  7'3 

/  =  16,000  —  70  X  -  =  16,000  —(70  X  ^^  -  —  )  =  10,400  Ibs. 
r  \  1.44      / 

The  maximum  fiber  stress  in  compression  being  10,940  Ibs.  the 
section  will  do. 

Purlins.  —  The  distance  between  trusses  being  17  ft.,  the  pur- 


lins will  be  considered  as  beams  supported  at  the  ends,  having 
a  span  of  17  ft.  and  carrying  a  uniform  load  of  35  Ibs.  per  sq.  ft., 
being  5  Ibs.  less  than  the  assumed  roof  load  on  the  truss  to  allow 
for  the  truss  weight  which  of  course  does  not  come  upon  the 

purlins.     The  distance  between  purlins  is  —  -  =  4.71  ft.     The 

load  carried  by  one  purlin  134.71  X  17  X  35  =  2800  Ibs. 

WT 

M  =  ~=  =  (2800  X  17  X  12)+  8  =  71,400  in.  Ibs. 
8 

/  _  M  _  71,400  _ 
e  ~  p  ~  16,000 

This  will  require  6-in.  channels  at  8  Ibs.  per  ft. 

Columns.  —  The  height  from  the  ground  to  the  top  of  the 
ventilator  being  48  ft.,  the  vertical  projection  between  adjacent 
columns  is  48  X  17  =  816  sq.  ft.  Taking  the  horizontal  wind 
pressure  at  15  Ibs.  per  sq.  ft.  of  vertical  projection  the  total 
pressure  on  the  building  between  adjacent  columns  is  816  X  15 
=  12,240  Ibs.  Assuming  the  horizontal  reactions  at  the  bases 
of  the  two  columns  in  any  transverse  bent  as  equal,  this  hori- 

zontal reaction  is  I2'24°  =  6120  Ibs. 
2 


DESIGN  OF  A   STEEL  MILL  BUILDING  127 

The  bending  moment  on  a  column,  assuming  that  the  knee 
braces  run  down  the  column  a  distance  of  5  ft.  making  the  dis- 
tance from  the  foot  of  the  knee  brace  to  the  base  of  the  column 
15  ft,  is 

M  =  6120  X  15  X  12  =  1,101,600  in.  Ibs. 

Trying  a  column  made  up  of  four  4  in.  X  3  in.  X  iVin.  angles 
and  one  16  in.  X  tVm-  plate  we  must  first  determine  the  area 
and  then  the  moment  of  inertia  of  the  section.  The  area  of  the 
section  is: 

4  angles,    4  X  2.09  =      8.36  sq.  ins. 

i  plate,      T\  X  16    =      5.00  sq.  ins. 

Total  area    13.36  sq.  ins. 

Moment  of  Inertia  of  Section.  — 

4  angles,  4X1.65  =      6.60 

Ah2  of  angles,  4  X  2.09  (8.25—0.76)  =  470.00 

W3      5        i63 
,-,  -X-  =    IQ7.oo 


/  =   583.60 

The  extreme  fiber  stress  in  the  column  due  to  bending  then 
is 

,      Me  8.25 

/  =  —  =  1,101,000  X        J  =  15,600  Ibs.  per  sq.  in. 
i  583-0 

The  direct  load  in  the  column  due  to  the  assumed  equivalent  load 
of  40  Ibs.  per  sq.  ft.  of  horizontal  projection  being  29,250  Ibs., 

the  fiber  stress  due  to  this  load  is  29'25°  =  2200  Ibs.  per  sq.  in. 

13-36 

The  total  maximum  fiber  stress  in  the  extreme  fibers  is  15,600 
-f  2  200  =  17,800  Ibs.  The  column  must  now  be  tested  by  the 
straight-line  or  other  column  formula  to  see  if  this  fiber  stress  is 

excessive.     Since  the   radius  of  gyration  equals  y  —  >  we  have 


13-36 
The  length  of  the  column  to  the  knee  brace  divided  by  the 

radius  of  gyration  is  15  X  —  =  27.3,  and  this  value  substituted 


128  GRAPHICS  AND  STRUCTURAL   DESIGN 

in  the  straight-line  formula  gives  the  allowable  stress  per  square 
inch  as 

/  =  1.25^16,000  -  yoX-  J  =  1.25  [16,000  -  (70  X  27.3)] 

=  17,600  Ibs. 

The  factor  1.25  is  used  here  in  accordance  with  usual  specifica- 
tions which  allow  an  increase  of  25  per  cent  in  the  working  fiber 
stress  when  the  stress  is  due  to  a  combination  of  wind,  snow  and 
dead  loads.  The  agreement  between  this  latter  stress  and  that 
found  as  existing  in  the  column  is  so  close  that  the  column  section 
tried  is  satisfactory. 

The  column  should  also  be  examined  for  its  strength  about  its 
other  axis.  The  inertia  about  axis  2-2  is  found  to  be  35.4.  The 
total  area  of  the  column  at  the  knee  brace  and  below  that  point 

is  i6.oosq.  ins.,  from  which  r  =  \  —•  =  1.49. 

A 

The  unsupported  height  being  8  ft., 

-  =  8  X  —  =  64.5. 
r  1-49 

This  value  of  -  being  within  the  permitted  limits  the  column  is 

satisfactory.  Where  the  lateral  wind  bracing  does  not  extend 
to  the  base  of  the  columns  the  reactions  due  to  the  wind  on  the 
end  of  the  building  should  be  divided  among  the  columns  and 
the  fiber  stress  in  the  column  due  to  the  resulting  bending  deter- 
mined. The  column  bases  may  be  considered  fixed  or  hinged, 
depending  upon  the  bases,  anchor  bolts  and  weight  of  foundation. 

WIND  BRACING 

The  reactions  at  the  tops  of  the  end  columns,  Fig.  161,  due 
to  wind  pressure  are 


2X3 


DESIGN  OF   A   STEEL   MILL   BUILDING 


129 


The  force  in  the  eave  strut  then  is  11,060  Ibs.  The  length  of 
this  strut  is  the  distance  between  columns,  or  17  ft. 

For  an  eave  strut  try  Fig.  162,  composed  of  four  3  in.  X  2\  in. 
X  J-in.  angles  latticed,  the  long  legs  of  the  angles  being  turned 
out, 

r  =  I7XZ^=I45; 

according  to  the  straight-line  formula 

/  =    16,000  —  (70  X  -J  X  1.25  =  7300  Ibs.  per  sq.  in. 

-J- 

'to 

~«O 


-SCO- 


FIG.  161. 


FIG. 


The  total  force  on  the  four  angles  is  4  X  1.32  X  7300  =  38,500  Ibs. 
No  lighter  section  can  be  used  as  these  are  the  lightest  allowed 

by  the  specifications,  for  the  -  value  determines  the  section. 

The  transverse  brace  between  the  first  and  second  trusses  will 
be  designed  as  a  simple  truss  to  carry  the  wind  loads  at  its 
several  apexes. 

Having  determined  the  apex  loads  in  Fig.  163,  the  stresses 
in  the  truss  members  can  be  found  either  by  the  method  of 
coefficients  or  by  making  a  stress  diagram,  Fig.  164.  The 
diagonals  will  be  assumed  as  taking  only  tension.  The  maximum 
stress  is  in  the  end  diagonal  and  is  found  to  be  13,800  Ibs. 
Trying  a  3  in.  X  2^  in.  X  J-in.  angle,  which  is  the  lightest 
allowed,  and  assuming  one  rivet  hole  at  a  section,  the  net  area 
is  1.63  —  0.50  =  1.13  sq,  ins.  The  total  load  it  will  carry, 


I30 


GRAPHICS  AND   STRUCTURAL  DESIGN 


allowing  a  fiber  stress  of  16,000  Ibs.  per  sq.  in.,  is  1.13  X 
16,000  =  18,100  Ibs.  As  the  stresses  in  the  diagonals  of  the 
braces  in  the  top  chords  of  the  roof  trusses  and  the  bracing 
between  the  columns  are  both  lower  than  the  stress  in  the  above 
diagonal  the  3  in.  X  2§  in.  X  i-in.  angle  will  be  used  in  all  these 
places.  The  strut  in  the  bracing  in  the  upper  chord  of  the 
trusses  will  be  stiffened  by  attaching  it  to  the  purlin.  Here  a 
3  in.  X  3  in.  X  |-in.  angle  will  be  used. 


FIG.  163. 


F  11100  # 

FIG.  164. 


GIRTS 


The  load  on  the  girts  134X17X15  =  1020  Ibs. 

M  =  ——  =  1020  x  17  X  —  =  20,800  in.  Ibs. 
o  o 


20,800 


=  1.30. 


/       e       16,000 

A  5  in.  X  3^  in.  X  -f^-in..  angle  will  be  used.  The  struts  in  the 
transverse  truss  will  be  made  of  two  5  in.  X  3^  in.  X  -rVm- 
angles,  with  the  long  legs  parallel.  The  radius  of  gyration  here 

I  I2 

being  1.46  the  value  of -is  17  X-—  =  140. 

r  1.46 

In  the  following  problem  (Fig.  165)  the  stresses  due  to  dead 
load,  snow  load  and  wind  load  have  been  tabulated  and  their 
resultants  compared  with  the  stresses  produced  by  an  equivalent 
dead  load  of  40  Ibs.  per  sq.  ft.  of  horizontal  projection  of  the 
roof. 

The  building  is  of  somewhat  lighter  design  than  the  previous 
one.  Span  of  truss,  86  ft.  o  ins.;  rise,  J  span.  Truss  spacing, 


DESIGN  OF  A  STEEL  MILL  BUILDING 


132  GRAPHICS  AND  STRUCTURAL  DESIGN 

25  ft.  o  ins.  center  to  center.  Figures  167  and  168  are  the  stress 
diagrams  for  wind  load  and  dead  load.  The  horizontal  wind 
pressure  is  assumed  at  20  Ibs.  per  sq.  ft.  and  the  normal  pres- 
sure on  the  roof  is  taken  as  that  given  by  the  formula 

PN  =  —  X  PH  =  ^  X  20  =  12  Ibs.  per  sq.  ft. 
45  45 

The  wind  is  shown  as  acting  upon  the  left  of  the  truss,  and  the 
truss  members  in  compression  are  represented  by  heavy  lines. 

The  dead-load  stress  diagram  is  given  in  Fig.  168  and  has  been 
made  to  serve  for  both  dead  load  and  snow  load  by  changing  the 
scale.  The  snow  load  has  been  assumed  at  20  Ibs.  per  sq.  ft.  of 
horizontal  projection.  The  weight  of  the  truss  is  taken  as 


300  — 

where 

W  =  weight  of  truss  per  square  foot  of  building. 
L  =  span  of  truss,  in  feet. 

D  —  distance  center  to  center  of  trusses,  in  feet. 
P  =  load  per  square  foot  on  truss. 
W  4Q  X  86 

300  +  (6  X  86)  +  ^^ 
o 

The  estimated  total  weight  of  one  truss  is  =  3X25X86  = 
6450  Ibs. 
The  weight  of  the  purlins  is  assumed  as  that  given  by 


45  /      4 

where 

Wi  =  weight  of  the  purlins  per  square  foot  of  building,  in  pounds. 
PI  =  load  per  square  foot  on  purlins,  in  pounds. 
D  —  distance  center  to  center  of  trusses,  in  feet. 


-  1.3.25 


45  4 


DESIGN   OF   A   STEEL   MILL   BUILDING  133 

The  roof  covering  will  be  No.  20  galvanized  corrugated  iron, 
weighing  about  2  Ibs.  per  sq.  ft.  or  2.75  Ibs.  allowing  for  laps, 
etc. 

The  wind  load  per  square  foot  of  horizontal  projection  is  the 
normal  wind  pressure  per  square  foot  multiplied  by  the  secant 
of  27  degrees,  the  angle  of  the  roof,  and  equals  12  X  1.12  =  13.44 
Ibs.  The  snow  load  will  be  20  Ibs.  per  sq.  ft.  of  horizontal 
projection. 

The  total  load  per  square  foot  of  horizontal  projection  is 

Truss  and  purlins  (3  +  3.25),  say 6 .  25 

Roof  covering 2.75 

Snow  load 20 .  oo 

Wind  load 13 .50 


Total 42 . 50 

It  is  commonly  specified  that  roofs  of  this  character  shall  be 
designed  for  a  uniform  dead  load  of  40  Ibs.  per  sq.  ft.  of  horizontal 
projection.  This  is  treated  as  an  equivalent  loading  and  replaces 
the  wind,  snow  and  dead  loads. 

The  stresses  will  now  be  determined  for  the  estimated  dead, 
snow  and  wind  loadings  and  their  resulting  maximum  stresses 
compared  with  stresses  determined  from  the  uniform  dead  load 
of  40  Ibs.  per  sq.  ft. 

The  apex  dead  load  on  the  upper  chord  of  the  truss  at  10  Ibs. 
per  sq.  ft.  is 

86  X  25  X  io 

-  =  1 790  Ibs. 

The  apex  snow  load  at  20  Ibs.  per  sq.  ft.  of  horizontal  projection 
is 

20  X  25  X  86 

—  =  3  580  Ibs. 

The  apex  normal  wind  load  at  1 2  Ibs.  per  sq.  ft.  of  roof  surface, 
the  distance  between  the  panel  points  along  the  upper  member 
of  the  truss  being  8.05  ft.,  is  8.05  X  25  X  12  =  2415  Ibs. 


134 


GRAPHICS  AND   STRUCTURAL  DESIGN 


Member. 

Dead  load  at 
10  Ibs. 

Snow  load  at 
20  Ibs. 

Wind  load  at 
12  Ibs.  (normal). 

Combined. 

Equivalent  load 
at  40  Ibs. 

CJ 

—  22,500 

—  45,000 

—  12,500 

—  8o,OOO 

—  90,000 

.  JK 

—    2,OOO 

—   4,000 

—    2,800 

—    8,800 

—  8,000 

LM 

+  3,6oo 

+  7,200 

+   5,300 

+  16,100 

+  14,400 

QT 

JA 

+  9.300 
+  20,500 

+18,600 

+41,000 

+  IO,6oo 
+  I7»900 

+38,700 
+  79,400 

+37,200 
+82,000 

AT 

+  11,000 

+  22,000 

+  5,ioo 

+38,100 

+44,000 

TU 

—  22,500 

-45,000 

—  13,800 

—  81,300 

—  90,000 

MN 

—  4,800 

—  9,600 

—  6,700 

—  21,100 

—  19,200 

The  selection  of  the  material  for  the  members  will  now  be 
made  (see  Fig.  169),  the  stresses  given  for  the  equivalent  load 
being  used. 

Lower  Chord.  —  The  maximum  stress  occurs  in  the  lower 
chord  in  the  piece  JA  (Fig.  165)  and  is  82,000  Ibs.  in  tension. 
Allowing  a  working  stress  of  16,000  Ibs.  per  sq.  in.  and  assuming 
f-in.  diameter  rivets  driven  in  yf-in.  holes,  we  find  the  net 


section  required  to  be 


82,000 


=  5. 13  sq.  ins. 


16,000 

Two  6  X  3^  X  -rVm-  angles  having  a  gross  area  of  5.78  sq.  ins. 
after  deducting  i  rivet  hole  in  each  angle,  that  is,  an  area  yf-in. 
long  and  yVm-  wide,  gives  a  net  area  of  5.78  —  (2  X  0.25)  = 
5.28  sq.  ins.,  which  is  sufficiently  close  to  the  required  net  area. 

As  the  trusses  will  have  to  be  divided  for  shipment  the  lower 
chord  member  AT  will  be  made  lighter  than  the  members  JA 
and  MA.  The  stress  in  AT  is  44,000  Ibs.  tension.  The  net 

area  required  is  4j"'00°    =    2.75  sq.  ins.     Trying  two  angles 
16,000 

3^  ins.  X  i\  ins.  X  -£$  in.  whose  gross  area  is  2  X  1.78  =  3.56 
sq.  ins.,  the  net  section  =  3.56  —  (2  X  0.25)  =  3.06  sq.  ins., 
which  is  sufficiently  close  to  the  required  2.75  sq.  ins. 

Other  Tension  Members.  —  The  members  NT  and  QT  will  be 
made  of  the  same  angles,  and  as  the  stress  in  QT  is  the  greater 
the  selection  will  have  to  be  made  for  it.  The  net  section 


16,000 


=  2.32  sq.  ins.      Trying  two  3^  in.  X  2\  in.  X  J-in. 


DESIGN  OF  A  STEEL   MILL  BUILDING 


135 


angles  whose  gross  area  is  2  X  1.44  =  2.88  sq.  ins.,  the  net  area 
becomes  2.88- (2  X  0.20)  =  2.48  sq.  ins.,  which  will  do. 


Eave 
L»3H 
8.L:1.L. 


J 

<3xjodk 


a.. 


2-  L  .  3^'x  2^  x - 


FIG.  169. 

The  members  LM  and  M)  have  approximately  the  same  stress. 
Taking  LM,  which  carries  14,400  Ibs.  in  tension,  the  net  area 


required  is  =  0.92  sq.   n. 

16,000 


136  GRAPHICS  AND   STRUCTURAL  DESIGN 

One  angle  2\  X  i\  X  J  in.  gives  a  net  area  of  1.19  —  0.20  = 
0.99  sq.  in.,  which  is  satisfactory. 

Compression  Members.  —  Member  MN  having  a  length  of 
ii  ft.  6  ins.  carries  19,200  Ibs.  in  compression.  If  the  maximum 
ratio  of  length  to  least  radius  of  gyration  is  125  then  the  least 
radius  of  gyration  is  length  divided  by  125  or  jf  f  =  i.io.  The 
smallest  angles  that  will  serve  are  3^  ins.  X  2  J  ins.  X  i  in.,  the  long 
legs  back  to  back,  and  separated  f  in.  Their  area  is  2.88  sq.  ins. 
According  to  the  straight-line  formula  the  allowable  stress  is 

p  =  16,000  —  70-  =  7250  Ibs. 

The  total  load  is  7250  X  2.88  =  20,900  Ibs.  This,  being 
slightly  in  excess  of  the  19,200  Ibs.  stress  in  the  member,  is 
satisfactory. 

Compression  members,  JK,  KL  and  the  corresponding  ones  on 
the  other  side  of  the  truss  have  the  same  load  of  8000  Ibs.  and 
are  approximately  84  ins.  long.  The  least  radius  of  gyration  is 
y82\  =  0.67.  The  smallest  angles  are  i\  ins.  X  2  ins.  X  J  in.,  being 
also  the  smallest  ordinarily  permitted  in  this  class  of  work.  That 
is,  no  leg  containing  rivets  shall  be  less  than  2§  ins.  and  no  material 
ihall  be  less  than  \  in.  thick.  The  minimum  radius  of  gyration  for 

the  angle  is  0.70  and  the  -  value  for  this  member  is  — —    =  106. 

r  0.79 

The  allowable  stress  is  p  =  16,000  —  70  -  =  8580  Ibs.     The  total 

load  on  the  section  is  2  X  1.07  X  8580  =  18,400  Ibs.  Although 
this  greatly  exceeds  the  load  on  the  member  it  is  the  smallest 
permitted  by  the  specifications  and  will  be  used. 

The  Upper  Chord  or  Rafter.  —  The  distance  between  apices 
along  this  chord  is  approximately  8  feet.  The  piece  is  subjected 
to  combined  compression  and  flexure.  It  will  be  assumed  that 
purlins  are  spaced  6  ft.  o  ins.  center  to  center  and  that  the  maxi- 
mum bending  will  occur  when  a  purlin  lies  midway  between  two 
adjacent  apices. 


DESIGN  OF   A   STEEL   MILL   BUILDING  137 

The  load  carried  by  each  purlin  is  6  X  25  X  34  =  5100  Ibs. 
Considering  the  continuity  and  the  method  of  securing  it  at  the 
points  of  support  it  is  usually  safe  to  take  the  bending  as  five- 
eighths  of  that  on  a  beam  similarly  loaded  but  supported  at  the 
ends.  The  bending  on  a  section  of  the  upper  chord  whose  span 
is  84  ins.  is 

M  =  ^X  —  =^X  (5100  X  84)  -*•  4  =  67,000  in.  Ibs. 

84^ 

The  maximum  compression  along  the  upper  chord  is  90,000  Ibs. 

and  occurs  in  CJ.    The  design  will  be  carried  out  for  this  mem- 

ber.    The  proper   section   can   only   be 

determined   by   trial,  that  is,  a  section         '=s 

must  be  assumed  and  then  tested  by  cal-    l       f      Inr— 

culation  to  see  if  it  satisfies  the  condi-          :j, 

tions.     The  section  most  frequently  used 

is  that  shown  in  Fig.  170.      The  plate  is        —  i__ 

generally  taken  of  sufficient  width  so  that  FlG 

the   members    joining   the   upper   chord 

may  be  riveted  directly  to  it.     The  section  to  be  tried  will  be 

made  up  of  two  4  in.  X  3  in.  X  iVin.  angles  and  one  10  X  iVin. 

plate. 

It  is  first  necessary  to  find  the  center  of  gravity. 

Area.  Statical  moment. 
Two  angles  2  X  2.09  =  4.18  X  0.76  =  3.18 

One  plate  10  X  0.313  =  3.13  X  5-0  =  I5-^5 

Total  area  =  7.31.        Total  moment  =  18.83 

*'=  Wi8-=  2.  58  ins. 


The  moment  of  inertia  about  axis  i-i  then  readily  follows: 

Inertia  of  the  two  angles  about  their  axis  2  X  1.65  =  3.30 

Ah*  of  these  angles  about  common  axis  4.18  X  (2.58  —  0.76)*  =  13  .90 

bd3      0.31  X  io3 

Inertia  of  plate.  —  =  —             -  =  26.00 
12              12 

Ah2  of  plate,  3.13  X  (5-00  -  2.58)2  =  18.30 

/  *=  61.50 


138  GRAPHICS  AND  STRUCTURAL  DESIGN 

Me 
The  fiber  stress  due  to  bending  is  given  by  /  =  •—  . 


fe  =  67,000  X  r~-  =  2800  Ibs.     Compressive  stress. 


L  =  67,000  X  *•   -  =  8080  Ibs.     Tensile  stress. 
61.5 

The  extreme  fiber  stress  at  backs  of  angles  is  —  12,300  —  2800  = 

—  15,100  Ibs. 

The  extreme  fiber  stress  at  edge  of  plate  is  —12,300  +  8080  = 

—  3920  Ibs. 

The  radius  of  gyration  of  the  section  about  the  axis  through 

the  center  line  of  the  plate  must  now  be  found.     Since  r  =  V/— 

v  A 

the  inertia  must  be  found  first. 

The  inertia  of  two  angles  about  the  axis  parallel  to  the  short 
legs  is 

2  x  3.38  6.76 

The  AW  portion  is  2  X  2.09  (1.26  +  o.i6)2  =  8.36 

Total  inertia  =»  15.12 

12 


1.44. 
o 


=  58-5,  say  60. 


/  =  16,000  —  70-  =  16,000  —  (70  X  60)  =  1  1,  800  Ibs. 

Since  wind  and  all  other  external  forces  have  been  considered  this 
allowable  fiber  stress  can  be  increased  25  per  cent,  making/  = 
14,800  Ibs.     This  being  within  5  per  cent  of  the  stress  estimated 
as  acting  in  the  piece,  the  15,100  Ibs.  fiber  stress  can  be  allowed. 

COLUMNS 

When  knee  braces  are  omitted  the  building  must  be  stiffened 
in  some  other  way.  This  may  be  done  by  designing  the  columns 
to  withstand  the  wind  loading  in  each  panel  on  the  side  of  the 
building,  by  designing  the  lateral  bracing  to  carry  all  wind 


DESIGN  OF  A   STEEL  MILL  BUILDING  139 

pressures  to  the  ends  or  by  assuming  a  division  of  this  loading 
between  the  columns  and  the  bracing.  Actually  the  columns 
and  bracing  always  share  such  loads,  but  as  the  proportion  carried 
by  each  depends  upon  their  relative  stiffness  the  actual  loading 
becomes  a  matter  of  considerable  uncertainty.  Just  what  as- 
sumptions are  best  to  make  will  depend  upon  the  length  and 
height  of  the  building,  the  assumed  wind  pressure,  the  manner 
of  securing  the  tops  of  the  columns  to  the  trusses  and  the  bases 
of  the  columns  to  the  foundations.  The  assumptions  for  the 
design  will  also  depend  upon  the  judgment  of  the  designer. 

All  bracing  should  be  given  an  initial  stress  when  erected,  to 
insure  its  acting  promptly  when  the  loading  comes  upon  it. 

The  portion  of  the  column  above  the  crane  girder  carries  the 
roof  load  and  will  be  liable  to  buckle  about  an  axis  parallel  to  the 
web.  The  load  is  applied  concentrically  to  the  column.  There 
may  be  bending  on  the  column  about  this  axis  from  wind  pressure 
on  the  end  of  the  building,  and  from  thrust  due  to  the  crane 
stopping  and  starting  on  its  runway  which  should  also  be  con- 
sidered. 

Below  the  crane  girder  the  inside  angles  of  the  column  must 
transfer  the  crane  load,  including  impact,  to  the  full  section  of 

the  column.     As  this  transfer  will  occur  in  a  short  distance  the  - 

r 

value  may  be  neglected,  hence  the  angles  on  this  side  of  the 

f  66,000 
column  would  require  a  minimum  section  of  — ==  4.13  sq. 

ins.  The  two  6  in.  X  3$  in.  X  iVm-  angles  used  give  an  area 
of  5.78  sq.  ins.  and  should  prove  ample. 

The  roof  load  and  the  side  of  the  building  carried  by  the 
columns  being  both  eccentric  to  the  axes  of  the  columns  will 
produce  bending  on  them. 

The  resultant  of  these  two  loads  and  the  crane  load  will  most 
likely  be  eccentric  to  the  center  of  gravity  of  the  columns  and 
will,  therefore,  produce  a  bending  moment  upon  them.  In 
addition  to  these  the  transverse  thrust  from  Jhe  crane  and  the 


140 


GRAPHICS  AND   STRUCTURAL  DESIGN 


wind  load  on  the  side  of  the  building  will  produce  bending  on  the 
columns  about  an  axis  at  right  angles  to  the  web.  The  effect  of 
these  moments  will  be  materially  reduced  by  the  restraint  at  the 
column  bases  if  the  columns  are  fixed  at  that  point.  The  maxi- 
mum fiber  stress  resulting  from  the  combination  of  direct  and 
flexural  stresses  should  not  exceed  that  permitted  upon  the 
column  when  the  maximum  allowable  stress  has  been  reduced 
by  a  suitable  column  formula.  Considering  that  the  stresses  are 

LOWER  CHORD  BRACING 


X 


END  VIEW 


^---^ 

~^^^ 

X 

Strut 

Eave 

Strut 

"XT 

X 

3><c 

SIDE  ELEVATION 
FIG.  171. 

the  resultant  of  wind  load,  crane  load  and  dead  load  this  reduced 
value  may  be  increased  25  per  cent.  The  column  section  below 
the  crane  girder  must  also  be  designed  for  a  possible  buckling 
about  an  axis  parallel  to  the  web,  and  for  bending  due  to  the 
crane  thrust  and  wind  load  on  the  end  of  the  building. 

The  reactions  at  the  tops  of  the  columns  are  assumed  as  carried 
to  the  ends  of  the  building  by  the  lateral  trusses  in  the  lower 
chords  of  the  roof  trusses.  As  members  of  the  lower  chords  of 
the  roof  trusses  form  parts  of  the  lateral  bracing  such  members 
should  be  examined  to  see  that  they  will  carry  these  bracing 
stresses  in  addition  to  those  already  on  them  from  the  truss  load. 
The  arrangement  of  the  bracing  is  shown  in  Fig.  171. 


CHAPTER    XI 
DESIGN    OF   A   RAILWAY   GIRDER 

Specification.  —  Design  the  girders  for  a  68-foot  span  deck- 
plate  girder  bridge  to  carry  a  single  track  for  Cooper's  E-6o 
loading,  given  in  the  table  on  page  71.  The  working  fiber  stress 
is  to  be  16,000  Ibs.  per  sq.  in.  in  tension  and  this  is  to  be  properly 
reduced  by  a  straight-line  formula  for  compression  members. 
The  allowable  shearing  fiber  stress  on  the  rivets  will  be  12,000  Ibs. 
per  sq.  in.,  while  24,000  Ibs.  per  sq.  in.  will  be  allowed  in  bearing. 
All  rivets  will  be  f  in.  in  diameter  and  rivet  holes  will  be  reamed 
to  i  in.  in  diameter. 

The  girders  will  be  placed  7  ft.  o  ins.  center  to  center.  The 
unit  shearing  fiber  stress  in  the  web  plate  shall  not  exceed  10,000 
Ibs.  per  sq.  in. 

The  dead  load  of  the  bridge,  two  girders,  shall  be  assumed  of 
the  following  weights  in  pounds  per  foot  of  span: 

Track 45° 

Steel,  girders,  bracing,  etc 1040 

Total 1490 

Allowable  pressure  between  base  plate  and  pier  shall  be  400  Ibs. 
per  sq.  in. 

Bending  Moments.  —  The  dead-load  bending  on  one  girder 

WL 

equals  -  -  at  the  center,  at  which  point  it  is  a  maximum. 

o 

M  =  IX      142°  x  68  X  68  X  ^  =  5,167,320^.  Ibs. 

o  2  o 

The  maximum  live-load  bending  moment  will  occur  when  the 
greatest  load  is  on  the  girder,  and  the  heavier  loads  near  the 
center.  The  bending  will  then  be  a  maximum  with  the  loads  so 

141 


142  GRAPHICS  AND  STRUCTURAL  DESIGN 

placed  on  the  girder  that  the  center  of  the  span  bisects  the  dis- 
tance between  the  center  of  gravity  of  the  loads  on  the  span  and 
the  adjacent  wheel.  The  bending  will  be  a  maximum  under  a 
wheel  and,  there  being  a  wheel  on  each  side  of  the  center  of 
gravity,  it  is  necessary  to  consider  each  of  the  two  wheels  in  turn 
as  the  adjacent  wheel  to  see  under  which  the  bending  moment  is 
the  greater.  Care  should  also  be  taken  to  see  if  any  of  the  loads 
in  the  assumed  loading  pass  from  the  girder  or  any  additional  ones 
come  on  it,  as  the  position  will  not  then  give  a  maximum  moment. 


-«-        ^ 

J 


4.07^ 


i r-f 

« M'0« J— *| 


FIG.  172. 

With  the  68-foot  span  under  consideration  to  have  heavy  wheel 
concentrations  near  the  center  when  the  bridge  is  fully  loaded,  it 
is  necessary  to  have  the  drivers  of  the  second  locomotive  in  the 
center  of  the  span.  Wheel  13  at  the  center  of  the  span  would 
bring  wheels  8  to  18  inclusive  on  the  span.  The  center  of  grav- 
ity of  this  group  must  now  be  found,  and  the  table  of  moments 
on  page  71  will  be  used  for  this  purpose.  The  moment  of 
wheels  8  to  18  about  wheel  18  from  the  table  is  7,525,500  ft.  Ibs. 
and  this  divided  by  the  sum  of  these  loads,  252,000  Ibs.,  gives 
29.86  ft.  as  the  distance  of  the  center  of  gravity  from  wheel  18. 
Now  placing  these  loads  on  the  span  in  accordance  with  the  pre- 
viously stated  conditions  for  maximum  bending  we  have  the 
loading  shown  in  Fig.  172. 

To  determine  the  bending  moment  we  must  first  find  the  left 
reaction.  Take  the  moment  of  the  loads  8  to  18  about  load  18 
and  add  to  it  the  moment  of  the  sum  of  the  loads  on  the  girder 
by  the  distance  the  right  pier  is  to  the  right  of  load  18.  The  sum 
of  these  moments  divided  by  the  span  will  give  the  left  reaction. 

R  =  [7.525,500  +  (252,000  X  4-07)]  -5-  68  =  125,750  Ibs. 


DESIGN  OF  A  RAILWAY   GIRDER  143 

The  bending  moment  under  load  13  is  then  equal  to  the  left 
reaction  multiplied  by  the  distance  from  the  left  pier  to  wheel  13 
less  the  moment  of  the  loads  on  the  girder  to  the  left  of  load  13. 
These  moments  also  are  taken  from  the  moment  table. 


or 


M  =  (125,750- X  33-93)  -  1,831,500  =  2,435,197  ft.  Ibs. 
M  =  2,435,200  X  12  =  29,222,400 in.  Ibs. 


The  moment  must  now  be  found  under  wheel  14  when  the 
center  of  the  span  bisects  the  distance  between  the  center  of 
gravity  of  the  loads  from  8  to  18  and  wheel  14.  Wheel  18  will 
then  be  6.57  ft.  from  the  right  pier  and  wheel  8  will  be  0.43  ft. 
from  the  left  pier.  As  the  same  loading  is  on  the  span  as  in  the 
previous  case  the  maximum  bending  moment  may  occur  under 
wheel  14.  The  moment  calculated  as  in  the  preceding  case  gives 
the  bending  under  wheel  14  as  28,407,360  in.  Ibs.,  which  is  less 
than  that  found  under  wheel  13  in  the  preceding  case. 

To  the  maximum  live-load  bending  must  be  added  the  dead- 
load  bending  and  the  bending  due  to  impact.  The  customary 
allowance  for  the  latter  varies  with  the  designer  but  will  here  be 
taken  as  that  given  by  the  formula 


300 

/  =  impact  to  be  added  to  the  live-load  strain. 
S  =  calculated  maximum  live-load  strain. 
L  =  length  of  loaded  distance  in  feet  which  produces  the 
maximum  strain  in  the  member. 

The  percentage  increase  in  the  live-load  stress  to  cover  impact  in 
this  problem  is 


=  8l.5  per  cent. 


The  maximum  impact  allowance  then  is  29,222,400  X  0.815 
23,816,256  in.  Ibs. 


144  GRAPHICS  AND  STRUCTURAL  DESIGN 

The  total  maximum  bending  moment  becomes: 

Dead-load  bending 5,167,320  in.  Ibs. 

Live-load  bending 29,222,400  in.  Ibs. 

Impact  allowance 23,816,260  in.  Ibs. 


58,205,980  in.  Ibs. 

In  designing  a  girder  it  is  usually  only  necessary  to  determine 
the  maximum  bending  moment,  so  the  above  method  is  all  that 
is  required.  If  it  is  desired  to  determine  the  bending  moments 
for  a  number  of  points  along  the  span  this  can  readily  be  done 
by  the  graphical  method  explained  on  page  47;  this  method 
has  been  used  to  determine  the  maximum  moment  and  is  shown 
in  Fig.  173.  This  diagram  gave  the  maximum  moment  as  oc- 
curring under  load  13  when  at  the  center  of  the  span,  and  the 
amount  of  the  bending  moment  was  estimated  at  29,400,000 
in.  Ibs.,  this  moment  differing  from  that  calculated  by  less  than 
i  per  cent.  The  calculation  gave  wheel  13  as  0.07  ft.  to  the  left 
of  the  center  of  the  span  which  is  in  very  close  accord  with  the 
diagram. 

Maximum  End  Shear.  —  A  diagram  of  maximum  shears  will 
be  drawn  (see  Fig.  175  and  explanation  on  page  49).  The  end 
shear  will  also  be  calculated.  The  maximum  shear  on  any 
section  will  generally  occur  when  the  span  to  the  right  of  that 
section  is  fully  loaded.  In  the  case  of  locomotive  wheel  loads, 
owing  to  the  considerable  difference  in  weight  between  the  pilot 
wheel  i  and  the  first  driver  2,  the  maximum  shear  may  occur  with 
wheel  2  at  the  section  rather  than  with  wheel  i  there.  In  this 
case  placing  wheel  2  over  the  left  pier  will  bring  wheel  13  on  the 
span  2  feet  from  the  right  pier.  The  reaction  then  is 

R  =  10>392,ooo  +  (318,000  ~  15.000)  X  2  =  I(5l?735lbs 
68 

The  diagram  of  maximum  shears  checks  this  value  of  the  end 
shear  and  shows  that  the  maximum  shear  occurs  with  wheel  2  at 
the  pier  instead  of  with  wheel  i  there. 


DESIGN   OF   A   RAILWAY   GIRDER 


DIAGRAMS  FOR  DECK  PLATE  GIRDER 
Span  68  ft.  Loading  Cooper's  E  60 


an%'njn K^ V  ftfr ?n p 


FIG. 


177. 


Wind  Bracing 
F  1  G  I   H  i    I 


&th  Web  plate-4.375/ 
—  Span68'0"—  »j 

Lengths  of  Flange  Plates 

FIG.  176. 


O     P 


u    v. 


FIG. 


8925^ 


x 


,17850^ 


End 
Bracing 


FIG.  179. 


T 

.1 


I'8'     MX* 


Stress  Diagram         / 

Q 

Kll\ 

/ 

/     \ 

/ 

R]\ 

/            \ 

/ 

u  7  \ 

/                    \ 

/ 

K/     \/            \/ 

C-J 


°'P 


FIG.  181. 


FIG.  1 80. 


146  GRAPHICS  AND  STRUCTURAL  DESIGN 

Dead-load  Shears.  —  The  weight  per  foot  of  one  girder  was 
previously  estimated  at  735  Ibs.  The  dead-load  end  shear,  being 
one-half  the  weight  of  the  girder,  is  equal  to  735  X  -28-  =  25,000 
Ibs.  The  dead-load  shear  varies  uniformly  across  the  girder 
changing  from  this  value  at  the  end  of  the  span  to  zero  at  the 

middle;  it  will,  therefore,  be-**1-   -  =  12,500  Ibs.  at  the  quarter 

point. 

The  impact  shear  will  be  given  by  the  formula  that  was  used 

for  impact  bending,  /  =  S I — ) .     This  gives  the  impact  as 

\L  +  3007 

81.5  per  cent  of  the  live -load  shear  when  the  girder  is  fully  loaded, 
and  the  impact  shear  becomes  0.815  X  161,735  =  131,810  Ibs. 
Combining  the  several  end  shears  we  have: 

End  shear,  dead  load 25>33°  Ibs. 

End  shear,  live  load 161,735  Ibs. 

End  shear,  impact 131,810  Ibs. 

318,875  Ibs. 

Allowing  a  unit  shearing  stress  of  10,000  Ibs.  per  sq.  in.  requires 
a  web  area  of  31.9  sq.  ins.     A  web  plate  80  ins.  deep  and  T7F  in.    - 
thick  has  an  area  of  35.0  sq.  ins.  and  is  ample. 

Flange  Area  and  Selection  of  Sections.  —  It  is  common 
practice  to  assume  the  effective  girder  depth,  that  is,  the  distance 
between  the  centers  of  gravity  of  the  girder  flanges,  as  approxi- 
mately the  distance  back  to  back  of  the  flange  angles.  After  the 
flange  sections  are  chosen  their  centers  of  gravity  are  determined 
and  this  distance  compared  with  the  distance  first  assumed;  if 
the  error  is  small  no  correction  is  made,  otherwise  the  distance 
is  taken  as  the  calculated  distance  between  the  centers  of  gravity 
of  the  flanges  and  new  flanges  are  determined  and  a  new  selection 
of  sections  is  made. 

The  girder  has  been  assumed  as  having  an  8o-in.  web  plate  and 
the  angles  will  be  set  out  \  in.  on  each  side  from  the  edge  of  the 
plate  to  avoid  the  possibility  of  the  plate  extending  beyond  the 
angles.  The  effective  depth  will  be  taken  as  80.5  ins.  The 


DESIGN  OF  A  RAILWAY  GIRDER  147 

allowable  working  fiber  stress  being  16,000  Ibs.  per  sq.  in.  the 
net  flange  area  is  found  by  dividing  the  maximum  bending 
moment  in  inch  pounds  by  the  product  of  the  allowable  fiber 
stress  and  the  girder  depth. 


Net  flange  area  =     58.205,980 
16,000  X  80.5 


45.2  sq.  ms. 


This  area  will  be  made  up  of  the  following  sections: 

One-eighth  web  area  =  i  X  35.0  =  4.4  sq.  ins. 

Two  6  X  6  X  f-in.  angles,  less  4  rivet  holes 13 . 9  sq.  ins. 

Three  14  X  f-in.  plates,  less  2  rivet  holes 27 .  o  sq.  ins. 

Total  area 45 .  3  sq.  ins. 

This  is  sufficiently  near  the  45.2  sq.  ins.  required.  The  rivet 
holes  have  been  assumed  as  i  in.  in  diameter.  As  the  calculated 
distance  between  the  centers  of  gravity  of  the  flanges  exceeds 
that  assumed  by  0.20  in.  the  section  assumed  will  be  slightly 
larger  than  required,  about  }  per  cent.  This  is  an  unimportant 
difference  and  will  be  neglected. 

The  live -load  shears  will  be  taken  from  the  diagram  of  maxi- 
mum shears,  Fig.  175,  and  to  these  shears  will  be  added  the  dead- 
load  and  impact  shears. 

Shears  in  pounds. 


End. 

1  span. 

\  span. 

Dead  load  

24,000 

12,495 

o 

Live  load 

162  ooo 

00,000 

46,000 

Impact  

131,800 

82,500 

40,400 

Total  end  shear  

318,700 

IQ3,QCK 

86,400 

At  the  quarter  point,  when  wheel  2  is  at  this  section,  the 
loading  extends  on  the  girder  59  ft.  from  the  right  pier  and 
according  to  the  formula  the  percentage  of  live-load  shear  to  be 
300 


allowed  for  impact  is 


=  83.5  per  cent.     The  impact  shear 


+  30° 

is  99,000  X  0.835  =  82, 500  Ibs.     In  the  same  way  the  allowance 
for  impact  at  the  center  of  the  span  is  found  to  be  40,400  Ibs. 


148  GRAPHICS  AND   STRUCTURAL  DESIGN 

Stiffening  Angles.  —  The  end  stiffeners  act  as  columns,  but  as 
the  load  is  shared  by  the  web  and  as  their  length  is  short  they  are 
generally  estimated  by  using  the  allowable  fiber  stress  but  not 
reducing  it  by  a  column  formula.  The  area  required  in  the  end 
stiffeners  then  is 

Max.  end  shear      318.875 

-  =     ,    -Lai  =  iQ-93  SQ-  ms- 
16,000  16,000 

Four  5  in.  X  3^  in.  X  rs~m-  angles,  having  a  combined  area  of  4  X 
5.38  =  21.52  sq.  ins.,  will  be  used.  The  intermediate  stiffeners 
are  not  usually  calculated.  Angles  of  the  same  dimensions  but 
of  lighter  section  are  used.  The  intermediate  stiffeners  will  be 
made  two  5  in.  X  3^  in.  X  f-in.  angles. 

The  four  end  stiffeners  being  over  the  end  bearing  plate,  the 
rivets  will  be  assumed  as  transferring  the  end  shear  from  the  web 
to  the  four  angles ;  that  is,  the  load  will  be  assumed  as  distributed 
over  the  entire  number  of  rivets  in  the  two  pairs  of  stiffeners. 
The  rivets  will  be  f  in.  in  diameter  and  will  be  in  double  shear 
and  will  bear  on  the  ^g-in.  we^  plate.  Allowing  a  shearing  fiber 
stress  of  12,000  Ibs.  per  sq.  in.  and  a  bearing  fiber  stress  of. 
24,000  Ibs.  per  sq.  in.  the  rivet  value  in  double  shear  is  14,430  Ibs. 
while  the  value  in  bearing  is  9190  Ibs.  The  number  of  rivets 

required  in  the  two  pairs  of  end  stiffeners  is  - — L-^  =  35- 

9190 

Lengths  of  Flange  Plates.  —  The  flange-plate  lengths  can  be 
approximated  as  in  Fig.  176.  Lay  off  a  line  representing  the 
span  of  the  girder  to  scale  and  at  its  center  erect  a  perpendicular 
representing  to  scale  the  calculated  net  flange  area,  in  this  case 
45.3  sq.  ins.  Now  upon  the  span  as  a  base  and  with  the  45.3  as 
an  altitude  draw  a  parabola.  The  parabola  construction  is  shown 
in  the  dotted  lines.  Starting  at  the  base  line  lay  off  in  succession 
distances  representing  one-eighth  the  web  area,  4.4  sq.  ins. ;  the 
net  area  of  the  two  flange  angles,  13.9  sq.  ins. ;  and  the  net  area 
of  the  three  flange  plates,  9  sq.  ins.  each.  The  distance  measured 
along  the  lower  line  of  any  section  between  the  sides  of  the 


DESIGN  OF  A  RAILWAY  GIRDER  149 

parabola  gives  the  theoretical  length  on  the  assumption  that 
the  bending  moment  across  the  girder  varies  as  the  ordinates  in 
the  parabola.  This  is  not  exact,  although  approximately  true. 
Were  greater  accuracy  desired  the  bending  moments  could  be 
estimated  for  intervals  along  the  girder  and  the  corresponding 
moment  diagram  drawn,  which  could  then  be  used  instead  of 
the  parabola. 

It  is  customary  to  take  lengths  of  plates  slightly  exceeding 
the  lengths  determined  in  the  way  just  described.  The  following 
are  the  scaled  lengths  and  the  lengths  used. 


Scaled  length. 

Actual  length. 

Outside  plate,  top  and  bottom 

Ft. 
3O 

Ft. 

•22 

Middle  plate,  top  and  bottom  ... 

A? 

46 

Inside  plate,  top.  .          

C2 

7O 

Inside  plate,  bottom  

C? 

C6 

Flange  Rivets.  —  First  consider  the  rivets  in  the  vertical  legs 
of  the  flange  angles  securing  these  angles  to  the  web  plate. 
The  rivets  transfer  the  change  in  horizontal  shear  from  the 
flanges  to  the  web  plate.  The  flange  forces  acting  at  intervals 
along  the  girder  can  be  determined  and  the  change  in  this  force 
between  adjacent  sections  used  to  estimate  the  number  of  rivets 
required.  The  change  in  flange  force  divided  by  the  rivet  value 
will  give  the  number  of  rivets  between  the  two  sections.  Be- 
sides this  change  in  horizontal  shear  the  rivets  also  transfer  the 
vertical  loads  to  the  web  through  the  upper  flange  rivets.  The 
maximum  wheel  concentrations  are  usually  considered  as  dis- 
tributed over  three  ties  or  approximately  36  ins.  along  the  flange. 

These  two  shears  can  be  combined  in  the  following  manner: 
Find  the  change  in  flange  force  per  inch  of  span  by  dividing 
the  difference  between  the  flange  forces  acting  at  two  adjacent 
sections  by  the  distance  between  these  sections;  this  will  give 
the  average  change  of  flange  force  per  inch  of  span  between  the 
sections  chosen.  The  vertical  shear  transferred  from  the  angles 


150  GRAPHICS  AND  STRUCTURAL  DESIGN 

to  the  web  plate  per  inch  of  span  will  be  the  maximum  wheel 
concentration  divided  by  36  ins.  The  square  root  of  the  sum  of 
the  squares  of  these  two  quantities  will  give  the  resultant  shear 
per  inch  of  span  within  the  given  section  and  the  rivet  value 
divided  by  this  resultant  shear  per  inch  will  give  the  rivet  spacing 
in  this  section. 

The  rivet  value  referred  to  is  the  lower  of  the  two  values  in 
bearing  or  shear. 

The  following  is  the  commonly  used  and  less  tedious  method 
of  determining  the  flange  riveting.  The  rate  of  change  in  flange 
force  per  inch  at  a  section  is  given  by 

f    v 
f    T 

where 

V  «=  maximum  vertical  shear  in  pounds  at  the  section. 
h  =  the  distance  between  rivet  rows  in  top  and  bottom 
flanges,   measured    in    inches.    Where    there    are 
double  rows  of   rivets  in  the  vertical   legs  of  the 
flange  angles  h  is  the  average  distance. 
Where  the  web  is  assumed  as  resisting  bending  the  value  / 
should  be  reduced  as  the  rivets   are  not  required  to  provide 
strength  to  secure  that  part  of  the  flange  to  the  web  which  is 
already  an  integral  part  of  it,  that  is,  the  one-eighth  web  area. 

A  =  total  net  area  of  flange,  and 
a  =  one-eighth  web  area,  then  the  reduced  force  becomes 

V      A-a 

*T?  ~T~ 

As  before  the  vertical  shear  per  inch  transferred  by  the  rivets 
due  to  the  maximum  wheel  concentration  is 
.       Max.  wheel  load 

/2=      ~W 

and  the  resultant  shear  per  inch  of  span  at  the  section  under 
consideration  is 


DESIGN  OF  A  RAILWAY  GIRDER  151 

The  rivet  spacing  then  is 

_  Rivet  value 

p'      /. 

The  rivet  spacing  is  generally  made  the  same  in  the  two  flanges. 

The  rivet  spacing  will  now  be  estimated  for  the  ends  of  the 
girder. 

The  average  distance  h  in  this  case  is  80.5  —  (3  X  2.25)  = 
73-75- 

/i  =  Fx.4-^==ai8I8oox  (4.4  +  I3-Q  +  9)-  44  =    6      lbs> 
h         A  73.75  4.4  +  13.9  +  9 

/,  =  V363o2  +  8352  =  3730  Ibs. 

The  rivet  value  in  bearing  for  f-in.  rivets  at  24,000  Ibs.  per 
sq.  in.  is  9190  and  the  rivet  spacing  then  is  p  =  f  £f  $  =  2.46  ins. 
Use  2j-in.  spacing. 

Rivet  spacing  a  distance  of  one-quarter  the  span  from  the  piers, 

fl  =  ZXA_I^=  194,000  X  31.9  =  ]bs 

*          A  73-75  X  36.3 

As  found  before, 

/2  =  835  Ibs. 

/3  =  V23io2  +  8352  =  2460  Ibs. 
P  =  fUnr  =  3-73  ins. 
Rivet  spacing  at  the  center, 

,       V  X  (A  -'a)      86,400  X  40-9 
/1=     -kXT       :  73-75  X  45-3   := 


/3  =      ioss2  +  8352  =  1350  Ibs. 
P  =  ttf*  =  6.8  ins. 

The  rivet  spacing  may  be  calculated  for  the  several  panels  by  the 
method  just  shown  or  it  will  be  given  with  sufficient  accuracy 
by  laying  off,  at  right  angles  to  a  line  representing  the  span, 
ordinates  showing  to  scale  the  rivet  spacing  at  the  ends,  the 


152  GRAPHICS  AND  STRUCTURAL  DESIGN 

quarter  points  and  the  middle  of  the  girder  and  then  connecting 
the  adjacent  points  with  lines.  The  spacing  at  any  section  may 
be  found  by  scaling  the  distance  from  the  axis  to  the  line  just 
drawn  at  the  particular  section. 

Riveting  of  the  Flange  Plates.  —  From  what  has  been  stated 
concerning  riveting  the  legs  of  the  flange  angles  to  the  web  plate 
it  will  be  evident  that  the  rivets  connecting  the  flange  plates  to 
the  angles  are  called  upon  to  do  much  less.  A  common  practice 
is  to  put  in  rivets  securing  the  plates  to  the  angles  and  have  them 
stagger  the  rivets  in  the  vertical  legs.  This  prevents  the  inter- 
ference of  the  rivets  and  furnishes  more  than  sufficient  rivets. 
Where  there  is  no  liability  of  interference  this  practice  need  not 
be  followed.  The  number  of  rivets  required  to  secure  a  plate  in 
place  and  develop  its  strength  can  be  estimated  by  multiplying 
the  net  area  of  the  plate  by  its  working  fiber  stress  and  dividing 
this  by  the  proper  rivet  value.  In  this  girder  the  plates  are 
14  X  |  in.  and  the  net  area  is  9  sq.  ins.;  at  16,000  Ibs.  per  sq.  in. 
the  force  in  the  plate  is  144,000  Ibs.  and  the  rivets  being  f  in. 
in  diameter  and  in  single  shear  their  value  is  7215  Ibs.,  which  is 
less  than  the  bearing  value  and  must,  therefore,  be  used.  The 
number  of  f-in.  rivets  in  single  shear  to  transmit  144,000  Ibs.  is 
144,000 


The  middle  plate  being  46  ft.  long  and  the  top  plate  33  ft.  long, 
the  middle  plate  extends  6.5  ft.  beyond  the  top  plate  at  each  end 
and  the  full  strength  of  this  middle  plate  must  be  developed  in 
6.5  ft. 

The  rivet  spacing,  considering  that  the  rivets  are  in  a  double 
row,  must  not  exceed 

6.5  X  12 

-  =  7.8  ins. 
10 

The  spacing  ordinarily  does  not  exceed  6  ins,  and  in  this  class 
of  work  is  frequently  specified  not  to  exceed  4  ins.  The  rivet 
spacing  is  also  commonly  made  less  at  the  ends  of  the  plates  and 
opposite  open  holes  left  for  field  rivets. 


DESIGN  OF   A  RAILWAY   GIRDER  153 

Wind  Bracing.*  -  -  The  wind  pressure  will  be  assumed  as  a 
horizontal  pressure  of  30  Ibs.  per  sq.  ft.  acting  upon  the  side 
of  the  girder  and  upon  a  train  assumed  as  having  an  average 
height  of  10  ft.  and  starting  2  ft.  6  ins.  above  the  rail.  The  wind 
load  will  be  assumed  as  transferred  to  the  ends  by  bracing  in  the 
plane  of  the  upper  flanges  of  the  truss.  At  the  ends  it  will  be 
transferred  from  this  bracing  to  the  piers  through  diagonal  cross 
bracing.  Lighter  intermediate  cross  frames  will  be  inserted  at 
intervals  along  the  span  to  act  as  spacers  and  to  stiffen  the 
girders,  but  these  are  not  usually  calculated.  Fig.  177  shows  a 
plan  of  this  bracing  in  the  plane  of  the  upper  flanges  and  Fig.  180 
is  the  stress  diagram.  The  apex  wind  loads  must  now  be  esti- 
mated. The  depth  of  the  girder  exposed  to  the  wind  pressure  is 
approximately  7  ft.,  which,  with  the  assumed  train  height,  makes 
a  total  of  17  ft. 

The  wind  load  per  foot  is  17  X  30  =  510  Ibs.     If  the  bracing 

is  divided  into  13  panels  the  apex  load  is  ^—      — *—  =  2750  Ibs. 

The  reaction  is  510  X  -/-  =  17,850  Ibs. 

The  diagonals  will  have  to  resist  both  tension  and  compression. 
The  maximum  stress  acts  in  piece  JK  whose  length  is  approxi- 
mately 8  ft.  6  ins.  The  stress  in  it  is  20,400  Ibs.  Allowing 

I  12 

-  =  1 20  the  smallest  radius  of  gyration  permitted  is  8.5  X  - 
r  1 20 

0.85. 

Trying  a  6  in.  X  4  in.  X  ^-in.  angle  the  value  of  -  =  —^  =  1 16. 

r      0.88 

The  allowable  stress  per  square  inch  is 
/=  16,000  —  70-  =/=  16,000  —  (70  X  116)  =  7880  Ibs. 

The  total  load  carried  by  the  6  in.  X  4  in.  X  J-in.  angle  is  4.75  X 
7880  =37,500  Ibs. 

*  Paragraph  144  of  the  specifications  would  make  the  loading  greater  than 
that  used  here. 


154  GRAPHICS  AND   STRUCTURAL  DESIGN 

To  develop  the  full  strength  of  this  piece,  using  f-in.  rivets  in 
single  shear  having  a  rivet  value  of  7200  Ibs.  and  allowing  25 
per  cent  extra,  as  the  rivets  are  field  driven,  would  require  37,500 

X^=7- 

720O 

The  maximum  stress  in  a  strut  will  be  that  in  the  end  frames 
which  will  be  assumed  as  one-half  the  reaction,  or  ''  ^  =  8925. 

Its  length  is  75  ins.     If  -  =  120,  r  =  -"•  =  0.625.     This  requires 

an  angle  not  less  than  3^  ins.  X  3^  ins.  and  one  angle  3!  ins.  X 
3!  ins.  X  |  in.  will  carry,  according  to  the  formula, 

/  =  16,000  —  70  -  =  16,000  —  (70  X  108)  =  8440  Ibs. 

The  total  load  is  2.49  X  8440  =  21,000  Ibs. 

The  number  of  f-in.  rivets  to  develop  the  strength  of  this 

section  as  a  column  is  21,000  X  ^^-  =  4. 

7200 

The  number  of  rivets  between  the  connecting  plate  and  the 
girder  may  be  determined  as  in  Fig.  179.  Lay  off  on  a  line 
parallel  to  PQ  a  distance  representing  the  number  of  rivets  in 
PQ  and  similarly  for  RS,  as  one  of  these  is  in  compression  and 
the  other  in  tension  their  components  along  the  main  girder  flange 
will  be  in  the  same  direction,  and  hence  the  number  of  rivets 
between  flange  and  plate  can  be  determined  by  projecting  these 
lengths  upon  the  horizontal  line  which  here  scales  10  rivets. 

End  Diagonals.  —  The  end  diagonals  are  7  ft.  8  ins.  long. 
When  one  is  in  tension  the  other  will  be  assumed  in  compression. 

The  load  carried  by  the  piece  is  12,600  Ibs.     If  -  is  not  to  exceed 

120  then  r  =  TVV  =  0.78.  6  in.  X  4  in.  X  f-in.  angles  will  meet 
this  requirement  and,  the  area  being  3.61  sq.  ins.,  is  more  than 
ample  to  carry  the  load. 

Web  Splice.  —  When  the  web  is  assumed  as  only  resisting  shear 
the  splice  is  calculated  to  provide  for  the  maximum  shear  at  the 


DESIGN  OF  A  RAILWAY   GIRDER  155 

spliced  section.  When,  however,  the  web  is  assumed  as  resisting 
bending,  that  is,  one-eighth  of  the  web  area  is  taken  as  acting 
at  the  flanges,  then  the  splice  plates  and  rivets  must  be  designed 
to  resist  a  corresponding  bending  moment.  The  depth  of  the 
splice  plate  will  be  the  distance  between  the  edges  of  the  verti- 
cal legs  of  the  flange  angles.  In  this  case,  allowing  ^  in.  for 
clearance,  the  depth  of  the  splice  plate  will  be  80.5  —  (12  -j-  ^)  = 
68  ins. 

The  fiber  stress  at  the  center  of  gravity  of  the  flanges  was 
assumed  as  16,000  Ibs.  per  sq.  in.  The  distance  from  the  neutral 
axis  of  the  girder  to  the  center  of  gravity  of  a  flange  was  assumed 

as  — —  =  40.25  ins.     The  fiber  stress  at  the  top  of  the  splice 

2 

plate  will  be  (16,000  X  V")  +  4°-25  =  I3j5°°  ^s.  per  sq.  in. 

The  bending  moment  resisted  by  the  web  is  —  -  X  depth 

8 

of  girder  X  fiber  stress  =  M  =  -\5-  X  80.5  X  16,000  =  5,635,000 
in.  Ibs.  The  area  of  the  splice  plate  required  to  resist  this 
moment  is  determined  by  a  calculation  similar  to  the  one  made 
for  the  web  plate.  This  area  is 

. 8  X  bending  moment  at  section,  in.  Ibs. 

depth  of  splice  plate,  ins.  X  extreme  fiber  stress,  Ibs.  per  sq.  in. 

8  X  5,635,000 

A  =  •  -  =  49.0  sq.  ins. 

68  X  13,500 

The  thickness  of  each  of  the  two  splice  plates  then  is 

/  =  — — —  =  0.36  in. 
2  X68 

If  the  rivet  farthest  from  the  neutral  axis  develops  its  full 
rivet  value  the  stress  upon  any  other  rivet  will  be  proportional 
to  its  distance  from  the  neutral  axis  and  the  moment  due  to  any 
rivet  being  also  proportional  to  this  distance  the  total  resisting 
moment  of  a  single  row  of  rivets  similarly  arranged  about  the 
neutral  axis  is 

M  =  2  X  rivet  value 


156 


GRAPHICS  AND   STRUCTURAL  DESIGN 


Here 


M  =  resisting  moment  in  inch  pounds  of  one  vertical  row  of 

rivets. 
R  =  rivet  value,  in  pounds.     (The  lower  of  the  values  in 

bearing  or  shear  should  be  used.) 

y  =  distance  of  the  rivet  from  the  neutral  axis,  in  inches. 
yi  =  distance  from  the  neutral  axis  to  rivet  in  splice  plate 

farthest  from  the  neutral  axis. 


Assuming  the  following  distances  of  the  rivets  in  one  row  above 
the  neutral  axis,  the  following  values  are  calculated  for 
Fig.  182. 


y2  in 


FIG.  182. 


FIG.  183. 


Number  of 
rivet. 

y 

y2 

Number  of 
rivet. 

y 

y- 

I 

o 

O 

7 

18 

324 

2 

3 

9 

8 

22.5 

506.25 

3 

6 

36 

9 

26.5 

702.25 

4 

9 

81 

10 

29-S 

870.25 

5 

12 

144 

ii 

32.5 

1056.25 

6 

15 

225 

3954-00 

The  rivet  value  of  a  f-in.  rivet  bearing  in  a  -^g-in.  plate  is 
9190  Ibs.,  hence  the  moment  of  one  row  of  rivets  spaced  as 
assumed  is 

2    ^^   O  T  OO 

M  =  -  -  X  3954  =  2, 240,000  in.  Ibs. 


DESIGN  OF   A  RAILWAY   GIRDER  157 

The  number  of  rows  required  on  this  basis  is 

15,635,000        i 

3'  OJ'     -  =  2\  rows. 

2,240,000 

This  would  take  three  rows  or  a  rearrangement  of  the  rivet 
spacing  to  increase  the  number  of  rivets  in  the  vertical  row  and 
at  the  same  time  the  moment  of  the  rivet  group. 

There  are  several  other  forms  of  splices,  one  of  which,  see  Fig. 
183,  consists  in  replacing  the  one-eighth  of  the  web  area  by  straps 
placed  on  each  side  of  the  flange  angles.  As  ordinarily  calculated 
the  area  of  the  strap  is  to  one-eighth  web  area  as  the  distance  back 
to  back  of  the  flange  angles  is  to  the  distance  a  center  to  center  of 
straps.  If  the  straps  are  9  ins.  wide  the  distance  center  to  center 
of  straps  will  be  59  ins. 

4.4  X  80.5 
Strap  area  =  -  —*  =  6.0  sq.  ins. 

59 
The  number  of  rivets  on  each  side  of  the  strap  will  be 

6.0  X  16,000 

-  =  10.4,  say  ii  rivets. 
9190 

The  central  plates  are  designed  to  resist  the  maximum  vertical 
shear  at  the  section. 

Still  another  method  is  to  splice  the  web  where  excess  material 
in  the  flange  makes  provision  for  all  or  part  of  the  bending  as- 
sumed as  taken  by  the  web  and  then  design  the  splice  for  the 
maximum  shear  and  any  bending  not  otherwise  provided  for 
occurring  at  the  section. 

In  the  problem  in  hand  the  web  splice  is  made  20  ft.  from  the 
piers,  and  the  shears  at  this  point  are  dead-load  shear  10,300  Ibs., 
live-load  shear  90,000  Ibs.,  and  the  impact  allowance  76,000  Ibs., 
making  a  total  shear  of  176,300  Ibs.  This  would  require  twenty 
f-in.  rivets  to  provide  for  the  maximum  shear  at  this  point. 
There  is  more  than  sufficient  surplus  material  in  the  flange  (see 
Fig.  176)  to  care  for  the  bending  assumed  as  coming  upon  the 
web,  so  that  a  splice  designed  for  shear  only  would  prove  ample. 


158 


GRAPHICS  AND   STRUCTURAL  DESIGN 


DESIGN  OF  A  RAILWAY   GIRDER  159 

Flange  Splices.  —  For  the  ordinary  girder  spans  splices  in  the 
angles  and  plates  of  the  flanges  are  not  quite  so  likely  to  be 
required  as  in  the  web.  When  such  splices  are  required  the  same 
general  principles  may  be  followed  as  in  the  case  of  the  web 
splice.  Splice  plates  may  be  placed  where  angles  or  plates  are 
cut  or  the  cut  may  be  made  where  the  flange  plate  would  other- 
wise end  and  then  the  splice  may  be  made  by  continuing  the 
flange  plate  beyond  the  cut.  This  affords  one  of  the  neatest 
ways  of  splicing  the  angles  or  flange  plates. 

Bed  Plates.  —  The  area  of  the  bed  plate  allowing  400  Ibs.  per 
sq.  in.  on  the  masonry  will  be  the  reaction  divided  by  400  or 

=  797  sq- ins- 

In  small  girders  carrying  light  loads  the  connection  between 
the  girder  and  the  masonry  may  be  made  by  the  use  of  a  couple 
of  rolled  plates.  These  should  extend  but  a  few  inches  beyond 
the  flange  plates  as  the  usual  depth  of  about  i  in.  is  not  stiff 
enough  to  transmit  much  load  to  the  pier  beyond  the  edges  of 
the  flange  plate.  The  bed  plate  rests  upon  the  masonry  while 
the  sole  plate  lies  on  the  bed  plate  and  carries  the  girder.  At 
one  end  the  girder  is  held  firmly  in  place  by  both  sole  and  bed 
plates  having  round  holes  a  little  larger  than  the  foundation 
bolts.  At  the  other  end  the  holes  in  the  sole  plates  are  slotted, 
permitting  the  movement  of  the  sole  plate  across  the  bed  plate 
due  to  the  expansion  and  contraction  of  the  girder.  The  plates 
at  the  moving  end  should  be  planed. 

Where  heavy  loads  are  carried  upon  girders  of  short  span, 
greater  depth  of  bed  plates  may  be  required;  this  type  of  cast- 
iron  bed  plate  is  shown  in  Fig.  188. 

In  the  longer  spans,  exceeding  75  to  80  feet,  greater  provision 
must  be  made  for  expansion  and  also  more  care  used  to  distribute 
the  load  on  the  masonry.  One  way  of  doing  this  is  shown  in 
Fig.  189.  Here  both  ends  of  each  girder  are  carried  on  pins, 
while  one  end  rests  on  friction  rollers.  The  pin  assists  the  distri- 
bution of  the  pressure  uniformly  over  the  rollers,  if  the  webs  of 


i6o 


GRAPHICS  AND  STRUCTURAL  DESIGN 


the  pedestal  are  of  ample  depth.  The  method  of  calculation 
would  be  as  follows:  Assuming  the  rollers  4  ins.  in  diameter  the 
allowable  load  per  inch  of  roller  length  is  L  =  1200  vQ,  where 
L  =  load  in  pounds  and  d  =  diameter  of  roller  in  inches.  L  = 
1 200  A/4  =  2400  Ibs. 

The  combined  lengths  of  all  the  rollers  in  the  nest  is  31  ^9°  = 

2400 

133  ins.     Making  the  rollers  23.5  ins.  long  would  require  -**•  ~ 

23-5 

6  rollers.    The  bearing  area  between  pin  and  web  is  - — *"?  = 

24,000 


_o       o       o      6       o       o       o 

, — =p= — r~ 


o       o       o      $       o 

0         0         0         0    !     0 

o      o 

ooo 

|33  e^fo-^-0-}^ 

5"f—  0  —  ^ 

0'Y0.)^ek>- 

231/3 en 


FIG.  189. 


13.3  sq.  ins.  If  three  i-in.  web  plates  are  used  the  projected  area 
is  3  ins.  X  I.OQ  in.  X  6  ins.  =  18  sq.  ins.,  and  the  area  will  be 
more  than  ample. 

The  web  plate  is  45  ins.  long  and  the  load  is  to  be  distributed 
over1  it.  A  rough  approximation  of  the  depth  of  the  web  from 
the  bottom  of  the  pin  to  the  lower  edge  of  the  web  plate  may  be 

made  by  considering  it  a  beam  section.    M  =  —^— = /  •  -  =  ^-~- ; 

o  c       6 


DESIGN  OF   A  RAILWAY  GIRDER  l6l 

here  d  =  depth  of  the  web  in  inches,  b  =  combined  thickness  of 
the  webs  in  inches. 

M     (-  ,72 

M  =  318,790  X-.r  =  16,000  X  3.00  X  —,  from  which  d  = 
8  6 

15  ins.  In  this  calculation  no  account  has  been  taken  of  the 
angles  and  plate  secured  to  the  lower  edge  of  the  web  plates,  so 
that  the  i5-in.  depth  used  should  prove  ample.  The  rollers  are 
kept  in  place  by  two  flats  held  by  three  rods  shown  at  the  ends 
and  the  middle.  Angles  at  the  sides  secured  to  the  bed  plate 
prevent  the  rollers  ,f rom  moving  laterally.  The  bending  on  the 
pin  is 

318,700  X  1.25 

* — LL2      2  =  132,500  in.  Ibs. 

•5 

On  the  basis  of  an  allowable  extreme  fiber  stress  in  pins  of 
22,000  Ibs.  per  sq.  in.  a  pin  6  ins.  in  diameter  will  resist  a  bend- 
ing moment  of  466,500  in.  Ibs.,  so  that  in  this  way  the  pin  is 
exceedingly  strong. 

The  methods  indicated  illustrate  a  couple  of  the  simplest 
methods  used  for  securing  girders  in  place. 


CHAPTER  XII 
CRANE    FRAMES 

FIGURE  190  represents  a  frame  for  an  underbraced  jib  crane. 
The  live  load  it  is  to  carry  is  3  tons  (6000  Ibs.).  To  this  must  be 
added  an  assumed  weight  of  block,  trolley,  chain,  etc.,  and  this 
will  be  taken  at  500  Ibs.  The  total  live  load  thus  becomes 
6500  Ibs.  The  dead  load  is  made  up  of  the  weight  of  the  frame 
and  hoisting  machinery  and  will  be  assumed  at  2000  Ibs.  The 
line  of  action  of  this  dead  load  will  be  taken  at  three-tenths  of 
the  effective  radius  from  the  mast,  or  11.75  X  0.3  =  3.5.  The 
equivalent  load  at  apex  i  will  be  the  load  that  would  produce  the 
same  amount  on  the  mast,  and  is  2000  X  T3Q-  =  600  Ibs. 

The  chain  pull  at  different  points  in  its  length  will  be  estimated 
upon  an  assumed  efficiency  of  each  sheave  of  97  per  cent,  thus 
above  the  jib  the  chain  pull  will  be  6500  -r-  (2  X  Q-972)  =  3460 
Ibs.,  while  parallel  to  the  mast  it  is  6500  -j-  (2  X  o.973)  =  3560 
Ibs.  The  pull  on  the  racking  chain  will  be  assumed  at  1000  Ibs. 

The  maximum  direct  stress  will  occur  in  the  bracing  when  the 
load  is  at  its  maximum  radius.  To  determine  the  direct  stresses 
graphically  the  apex  loads  must  be  found.  To  calculate  the 
equivalent  apex  load  at  i  (Fig.  191),  when  the  load  is  at  its 
maximum  radius,  take  moments  about  point  3.  The  equivalent 

apex  load  at  i  =  6500  X  "'    '  =  7650  Ibs. 

10 

The  upward  reaction  at  point  3  is  7650  —  6500  =  1150  Ibs. 

The  diagram  (Fig.  192)  gives  the  combined  live- and  dead-load 
stresses;  these  can  be  determined  with  greater  accuracy  if  done 
separately  but  this  diagram  is  sufficiently  accurate  in  this  instance. 

The  jib  will  have  to  be  considered  for  bending  both  when  the 
load  is  at  its  maximum  radius  and  when  the  trolley  is  between 

162 


CRANE  FRAMES 


163 


FIG.  190. 


1 


Hbfeting  Chain 

^£^\       Racking  Chain "*~t  ~3 


FIG.  193. 


FIG.  194. 


FIG.  195. 


164  GRAPHICS  AND   STRUCTURAL  DESIGN 

points  i  and  2.  Here  the  span  is  60  ins.  and  the  distance  center 
to  center  of  the  trolley  wheels  is  taken  as  24  ins.  Although 
commonly  discussed  as  a  central  load  the  position  of  maximum 
bending  will  here  be  assumed  accurately ;  that  is,  when  a  wheel  is 
one-quarter  the  distance  between  the  wheels  from  the  center  of 
the  span.  Here  this  distance  is  \4-  =  6  ins.  (see  Fig.  193).  The 
load  of  6500  Ibs.  being  carried  upon  four  wheels  the  load  on  two 
wheels  or  an  axle  is  ^~  =  325°  IDS-  The  reaction 
R>  =  to°  X  24)  +  to°  X  48)  _  ,bs 

DO 

RI  =  6500  —  3900  =  2600  Ibs. 

The  direct  stress  in  the  section  of  the  jib  carrying  the  trolley  is 
given  by  Fig.  194  as  6600  Ibs. 

Fiber  Stresses.  —  Crane  frames  are  liable  to  considerable 
shock.  The  usual  method  of  caring  for  this  is  to  take  a  lower 
working  fiber  stress  than  is  ordinarily  used  in  structures  not  sub- 
jected to  shock.  These  fiber  stresses  run  from  10,000  Ibs.  to  12,000 
Ibs.  per  sq.  in.  for  mild  steel,  which  is  the  material  commonly  used 
for  crane  frames.  The  allowable  stress  in  columns  is  the  above 
properly  reduced  by  a  suitable  column  formula.  As  it  is  difficult, 
if  not  impossible,  to  stiffen  some  parts  of  crane  frames  —  the  jib, 
for  instance,  in  the  present  problem  —  the  allowable  stress  in 
members  acting  as  beams  must  be  reduced  to  prevent  buckling 
of  the  compression  flange;  see  curves,  page  100. 

Selection  of  Jib.  —  The  load  at  the  maximum  radius  creates 
a  direct  stress  in  the  jib  of  13,900  lbs.rin  tension.  The  hoisting- 
chain  pull  is  3640  Ibs.,  while  the  pull  in  the  racking  chain  is 
1000  Ibs. ;  this  creates  a  force  of  1000  Ibs.  acting  in  the  jib  when 
the  trolley  is  pulled  back  and  of  2000  Ibs.  when  the  trolley  is 
drawn  forward.  The  average  height  of  the  two  racking  chains 
from  the  center  of  the  jib  is  taken  as  8  ins.  Calling  tension  +, 
and  compression  — ,  we  have : 

Direct  stresses  in  jib. 

Stress  due  to  live  and  dead  loads +  13,900  Ibs. 

Stress  due  to  hoisting  rope —    3,460  Ibs. 

Stress  due  to  racking  rope —    2,000  Ibs. 


CRANE   FRAMES  165 

The  greatest  stress  then  is  the  combination  of  the  stress  due  to 
live  and  dead  loads  with  the  stress  due  to  hoisting.  The  racking 
force  will  not  be  considered  as  it  gives  a  lower  stress  than  the 
above  combination.  The  maximum  stress  is  13,900  —  3460  = 
10,440  Ibs.  in  tension. 

The  bending  moment  is  (6500  X  23)  •  (3460  X  14)  = 
101,060  in.  Ibs.  The  stress  must  now  be  considered  when  the 
trolley  is  between  points  i  and  2.  The  maximum  .bending  = 
RI  X  24  =  2600  X  24  =  62,400  in.  Ibs.,  due  to  the  hook  load. 
To  this  must  be  added  the  bending  due  to  the  hoisting-  and  rack- 
ing-rope pulls.  The  former  is  M2  =  3460  X  14  =  48,400  in.  Ibs. 
That  due  to  the  latter  is  MS  =  2000  X  8  =  16,000  in.  Ibs.  The 
total  bending  is  62,400  +  48,400  +  16,000  =  126,800  in.  Ibs. 
The  direct  stress  is  6600  —  3460  —  2000  =  1140  Ibs. 

The  influence  of  span  to  flange  width  must  now  be  considered. 
Trying  8-in.  channels  at  n|  Ibs.  per  foot  the  flange  width  is 
2.26  ins.,  and  the  span  being  60  ins.  it  follows  that 

Span  ___  60         ^ 
Flange  width      2.26 

On  the  cantilevered  section, 

Span  23 

-  =  —  *-  =  10.2. 
Flange  width      2.26 

From  the  curve  the  allowable  stress  should  not  exceed  90  per  cent 
of  the  maximum  desired,  so  that  allowing  a  maximum  working 
fiber  stress  of  10,500  Ibs.  per  sq.  in.  it  reduces  to  10,500  X  0.90 
=  9450  Ibs.  per  sq.  in. 
The  section  modulus  for  an  8-in.  channel  at  nj  Ibs.  is  8.1; 

hence,  since  M  =  f-  and  the  bending  has  been  calculated  for  two 
£ 

channels,  we  have 

.      Me      126,800  ^,  0 
/=  —  =-         -X8.i  =  7850  Ibs. 


The  direct  stress  per  square  inch  =  ^--  X  3.35  =  170  Ibs.     The 
8-in.  channels  at  nj  Ibs.  will  be  satisfactory. 


1 66  GRAPHICS  AND   STRUCTURAL  DESIGN 

Member  AH.  —  The  length  of  this  piece  is  7  ft.  and  the 
force  acting  in  it  is  16,100  Ibs.  This  member  is  in  compression 

and  as  it  cannot  be  braced  and  its  -  value  must  be  limited  to  140 

r 

the  least  radius  of  gyration  will  have  to  equal  or  exceed  •££$  = 
0.60.  Seven-inch  channels  will  be  required  and  their  radius  of 
gyration  is  0.59.  According  to  the  abridged  form  of  Ritter's 
formula  for  soft  or  mild  steel 

f- L =      I0'5°°      =  IO-5°°  =  3550  Ibs 

i       m2         19,600      2.96 

i-l —        ~  XI  -  1        I~i — 

10,000      \rl  10,000 

and  the  total  allowable  load  on  two  7 -in.  channels  at  9!  Ibs.  per 
foot  or  2.85  sq.  ins.  in  the  section  is  3550  X  2  X  2.85  =  20,300 
Ibs.,  which  is  satisfactory. 

Member  AF.  —  This  is  a  compression  piece  carrying  a  load 
of  17,700  Ibs.;  its  length  is  approximately  13  ft.,  but,  as  it  can  be 
braced  across,  its  greatest  length  need  not  exceed  the  length  of 
the  piece  HA,  and  as  the  force  acting  in  it  is  about  the  same  as 
in  the  preceding  piece  it  will  be  made  of  the  same  section,  7-in. 
channels  at  9!  Ibs.  per  ft. 

Member  FG.  —  This  is  also  a  compression  piece;  its  length 
is  60  ins.  and  the  force  acting  in  it  is  10,500  Ibs.  Assuming  as 

before  that  the  -  value  shall  not  exceed  140,  r  must  not  be  less 
than  T6?°^  =  0.43.  Trying  5-in.  channels  at  6|  Ibs.,  which  have 

r  =  0.50,  we  find-  =  - —  =  120.     Substituting,  as  before,  in  the 
r      0.50 

column  formula  gives 

10,500  10,500  ,, 

=  4300  Ibs. 


xr 


10,000      \r/ 

The  total  allowable  force  on  the  piece  is  4300  X  2  X  1.95  = 
16,700  Ibs.,  and  these  channels  will  be  used. 

Mast.  —  For  most  positions  of  the  trolley  the  mast  will  be  in 
tension.  The  position  of  maximum  compression  will  require  the 


CRANE   FRAMES 


167 


The  sever- 


trolley to  be  placed  as  close  to  the  mast  as  possible. 
est  stress  on  the  mast  will  be  due  to  the  bending. 

The  horizontal  reaction  at  the  upper  pin  can  be  found  by  taking 
moments  about  the  lower  pin. 

R  X  13-75  =  (65°°  X  11.75)  +  K2000  X  11.75)  •*•  4]  =  6000  Ibs. 
The  bending  at  the  upper  portion  of  the  mast  equals  6000  X 
22  =  132,000  in.  Ibs.     The  mast  also  resists  bending  due  to  the 
5-in.  channels  which  fall  below  the  line  of  intersection  of  the  other 

! 
[tl 


Tin 


FIG.  198. 

pieces  at  apex  3.     The  reaction  on  the  mast  due  to  this  force  in 
FG,  the  horizontal  component  of  which  is  6800  Ibs.,  is 


RI  — 


6800  X  ii 


12 


=  6230  Ibs. 


and  the  bending  is 

M  =  6230  X  12  =  74,760  in.  Ibs. 

This  bending  is  in  the  opposite  sense  to  that  due  to  the  force  on 
the  pin. 

The  bending  due  to  the  hoisting-rope  pull  must  now  be  deter- 
mined. The  horizontal  reaction  at  apex  3  due  to  the  rope  pull  on 
the  drum  is  found  by  taking  moments  about  apex  5,  from  which 


l68  GRAPHICS  AND  STRUCTURAL  DESIGN 

The  maximum  bending  due  to  the  rope  pull  then  is 
M  =  173  X  90  =  15,57°  in-  tt>s. 

O-  If  fl  I  M  132,000 

Since      M  =  f  ->  -  =  —  =  -^— ^ =  12. z. 

e  e      f        10,500 

This  is  for  two  channels,  or  6.25  for  one,  and  would  require  7-in. 
channels  were  there  no  direct  stress.  Try  8-in.  channels,  the 
area  of  two  such  channels  being  2  X  3.25  =  7.0  sq.  ins.  The 
direct  stress  previously  found  was  9100  Ibs.  and  the  unit  fiber 

stress  is =  1300  Ibs.  per  sq.  in.     The  fiber  stress  allowed  for 

7.0 

bending  then  is  10,500  —  1300  =  9200  Ibs.  per  sq.  in.  It  follows 
that 

I      M      132,000 

-e  =  j-  -%£-  =I4-3S- 

This  will  take  two  8-in.  channels  at  nj  Ibs.  whose  section 
modulus  is  2  X  8.1  =  16.2. 

The  resulting  moment  at  any  section  is  the  intercept  between 
the  heavy  lines  in  Fig.  198. 


DESIGN  OF  FRAME  FOR  TOP-BRACED  JIB  CRANE 

The  capacity  of  the  crane  is  to  be  5  tons  (10,000),  at  a  maximum 
radius  of  20  ft.  The  weight  of  the  trolley,  hook  and  chain  will  be 
assumed  at  600  Ibs.  The  dead  load  of  the  frame  and  machinery 
will  be  taken  as  3000  Ibs.,  and  its  center  of  gravity,  or  line  of 
action,  will  be  considered  as  one-fourth  of  the  crane  radius  from 
the  mast,  or  %£-  =  5  ft.  The  load  which,  acting  at  apex  i,  would 
produce  the  same  moment  on  the  mast  as  the  frame  weight  is 

8JHLO    =    75olbs> 

The  maximum  fiber  stress  in  tension  is  to  be  12,000  Ibs.  per 
sq.  in.  In  compression  the  maximum  fiber  stress  will  be  12,000 
Ibs.  per  sq.  in.,  properly  reduced  by  a  column  formula.  The 

value  of  -  will  be  limited  to  140. 


CRANE   FRAMES 


169 


J 


The  skeleton  of  the  frame  is  shown  in  Fig.  199,  vwhile  the 
stress  diagram  is  represented  by  Fig.  201.  The  total  equivalent 
load  at  apex  i  is  the  sum  of  the  live  load,  10,000  Ibs.,  the  weight 
of  the  trolley,  hook  and  chain,  600  Ibs.,  and  the  equivalent  frame 


1 70  GRAPHICS  AND   STRUCTURAL  DESIGN 

load,  750  Ibs.,  making  a  total  of  11,350  Ibs.  The  horizontal  re- 
actions at  points  4  and  6  can  then  be  found  by  taking  moments 
about  apex  6. 

R  =  TI>35°  X  2°  =  12,845  Ibs. 
17-75 

These  several  external  forces  can  now  be  located  at  the  proper 
points  and  the  stress  diagram  drawn  in  accordance  with  the 
principles  given  in  Chapter  II. 

Selection  of  Members.  —  The  members  CE  and  H B  are  in 
tension  and  having  about  the  same  stress  acting  in  them  will  be 

£  A    OOO 

made  of  the  same  section.     The  net  area  required  is  — 

12,000 

4.50  sq.  ins. 

Trying  two  4  in.  X  3  in.  X  yVm-  angles,  and  assuming  |-in. 
diameter  holes  for  f-in.  rivets,  the  net  area  afforded  by  these 
angles  is 

Net  area  of  two  4  in.  X  3  in.  X  ^-in.  angles  =  (2  X  2.88)  —  (2  X 
0.38)  =  5.00  sq.  ins.,  which  is  satisfactory. 

The  member  CF  will  also  be  made  of  the  same  section,  as  it 
is  very  short. 

Member  GH.  —  The  maximum  stress  for  this  member  occurs 
when  the  load  is  at  apex  2.  The  stress,  given  by  Fig.  202,  is 

2*7  OOO 

27,000  Ibs.     The  net  area  required  then  is  -- =  2.25  sq.  ins. 

Trying  3  in.X  3  in.  X  i-in.  angles,  two  of  which  have  an  area  of 
2  X  1.44  =  2.88  sq.  ins.,  and  allowing  one  f-in.  diameter  hole  for 
a  f-in.  diameter  rivet  to  an  angle  the  net  section  for  two  angles 
is  2.88  —  (2  X  0.22)  =  2.44  sq.  ins.,  and  these  angles  will  do. 

Members  A  G  and  AH.  —  These  members  will  be  subjected  to 
combined  compression  and  flexure,  and  as  it  is  difficult  to  brace 
these  members  laterally,  due  consideration  must  be  given  to  the 
reduction  of  the  fiber  stress  to  provide  for  the  column  action  of 

the  compression  flange.  The  axle  load  is  -  >2-^-  =5325  Ibs. 
The  maximum  bending  on  the  lo-ft.  span  will  occur  when  one  of 


CRANE   FRAMES  171 

the  wheels  is  one-fourth  the  distance  between  the  wheels  from 
the  center  of  the  span. 

Finding  the  reactions  we  have 

R>  =  (53*5  X  54)  +  (5.^5  X  78)  =  ^ 

120 

Then  RI  =  10,650  —  5860  =  4790  Ibs. 

The  maximum  bending  moment  on  the  span  then  is 

M  =  RI  X  54  =  4790  X  54  =  258,660  in.  Ibs. 

The  maximum  stress  will  occur  in  these  pieces  in  member  HA 
when  the  load  is  located  as  shown  in  Fig.  204,  and  the  direct 


o    o 


— 


FIG.  204. 

stress  is  given  by  Fig.  203.  To  determine  this  stress  the  reaction 
at  apex  i  must  be  found  when  the  load  is  in  the  above  position. 

Under  this  loading  the  apex  load  at  i  is  R%  =  5860  Ibs.  plus  the 
equivalent  frame  load  at  i,  which  is  750  Ibs.,  these  making  a  total 
of  6610  Ibs. 

The  direct  stress  is  found  to  be  31,000  Ibs. 

The  jib  will  be  assumed  as  braced  in  two  places,  so  that  it  will 
be  approximately  braced  laterally  at  intervals  of  about  8  ft. 

This  bracing  can  be  only  a  light  frame,  generally  made  of  a 
bent  angle  that  must  clear  the  trolley.  The  total  bending  on 
the  member  AH  is  due  to  the  vertical  loading  used  in  deter- 
mining the  258,660  in.  Ibs.  just  calculated,  and  in  addition  to 
this  there  is  bending  due  to  the  hoisting-rope  pull  and  the  rack- 
ing-rope pull,  both  of  which  act  to  produce  bending  of  the  same 
character  on  the  beam  and  must  therefore  be  added  to  the  bend- 
ing moment  just  found. 


172  GRAPHICS  AND  STRUCTURAL  DESIGN 

The  bending  moment  due  to  hoisting-rope  pull  =  5650  X  19 
=  107,350  in.  Ibs. 

The  bending  moment  due  to  the  racking-rope  pull  =  1000  X  9 
=  9000  in.  Ibs. 

The  total  bending  moment  then  is  258,660  +  107,350  -f  9000 
=  375,010  in.  Ibs.  Assuming  that  a  section  will  be  tried  whose 
flange  width  is  3  ins.,  the  ratio  of  unsupported  part  of  span  to 

a             -j^  -    8  X  12 
flange  width  is  -  =32. 

From  the  curve  on  page  100  the  allowable  fiber  stress  in  com- 
pression for  this  ratio  of  length  to  flange  width  is  about  85  per 
cent  of  the  maximum  desired.  The  allowable  fiber  stress  is 
12,000  X  0.85  =  10,200  Ibs.  per  sq.  in.  If  we  try  i2-in.  channels 
at  25  Ibs.  per  ft.,  we  find  their  area  to  be  2  X  7.35  =  14.7  sq.  ins. 
and  their  section  modulus  2  X  24.0  =  48.0. 

The  direct  stress  per  square  inch  is  "— 2 =  2100  Ibs. 

14-7 

The  stress  permitted  for  flexure  becomes  10,200  —  2100 
=  8100  Ibs.  per  sq.  in.  The  required  section  modulus  then  is 

I  _  M_  _  375,010  _     . 
~e~~f=      8100 

It  is  evident,  therefore,  that  the  channels  tried  are  satisfactory. 
Before  being  accepted  finally,  however,  these  channels  should  be 
examined  as  columns  when  the  load  is  at  its  maximum  radius. 
Under  this  condition  the  jib  is  subjected  to  a  direct  stress  of 

c<2  tQO 

52,500  Ibs.     The  unit  stress  due  to  this  direct  load  is  °  ° 

14.7 

=  3570  Ibs. 
The  allowable  unit  stress  according  to  Ritter's  formula  is 

, 12,000 -  _    12,000    =  12,000 

'==        ^      -44 


io,ooo     \r/  10,000 

=  4900  Ibs.  per  sq.  in. 


CRANE   FRAMES  173 

The  -  value  of  120  was  found  by  dividing  the  unsupported 
length,  96  ins.,  by  the  radius  of  gyration  of  the  channels  about 


their  minor  axis,  0.78,  or  —  -  =  120. 

0.78 

The  i2-in.  channels  were  found  satisfactory  in  this  respect. 
Mast.  —  The  direct  stress  as  given  by  the  diagram,  Fig.  201, 
is  50,000  Ibs.     The  reactions  at  the  pins  are 


„       11,^50  X  20 
R  =  —  -^—        -  =  ii,  120  Ibs. 
20.5 

The  bending  moment  due  to  the  above  reaction  is 

M  =  n,  1  20  X  21.5  =  239,080  in.  Ibs. 
Trying   two   i2-in.  channels  at  20.5  Ibs.  per  ft.,  their  area  is 

2  X  6.03  =  12.06  sq.  ins.,  and  the  direct  compression  is^—i  — 

12.06 

=  4150  Ibs.  per  sq.  in.     Assuming  the  allowable  ultimate  fiber 
stress.  in  bending  as  12,000  Ibs.  per  sq.  in.  the  balance  for  bending 
is  12,000  —  4150  =  7850  Ibs. 
The  required  section  modulus  is 

I  =  M  _  239,080  _ 

~e      f  :       7850 

An  inspection  of  the  tables  of  the  properties  of  channels  in  a 
Manufacturer's  handbook  or  on  page  16  will  show  that  either 
a  lo-in.  channel  at  20  Ibs.  or  i2-in.  channels  at  20^  Ibs.  will  carry 
this  load.  However,  since  i2-in.  channels  are  used  for  the 
member  AG  and  as  these  channels  give  considerably  greater 
stiffness  they  will  be  used  rather  than  increase  the  number  of 
sections  demanded.  As  it  is  possible  to  brace  the  channels 
forming  the  mast  at  as  frequent  intervals  as  desired  no  account 
has  been  taken  of  the  ratio  of  span  to  flange  width. 


CHAPTER   XIII 

GIRDERS   FOR    OVERHEAD    ELECTRIC    TRAVELING 

CRANES 

Specification.  —  Design  a  bridge  for  a  lo-ton  O.E.T.  crane 
whose  span  is  45  ft.  Assume  the  distance  center  to  center  of 
trolley  wheels  as  4  ft.  o  ins.,  that  the  bridge  weighs  13,000  Ibs. 
(two  girders) ,  and  that  the  trolley  weight  is  6000  Ibs.  Design  the 
girders  to  resist  an  additional  loading  laterally  of  one-tenth  the 
crane  capacity.  Allow  a  working  fiber  stress  in  tension  of  10,000 
Ibs.  per  sq.  in.  and  the  same  properly  reduced  by  a  column 
formula  for  compression  pieces.  Use  f-in.  rivets  throughout, 
for  which  assume  f-in.  diameter  holes. 

The  loading  will  be  reduced  to  that  for  a  single  girder  or  one- 
half  the  bridge.  The  weight  of  the  bridge  will  be  assumed  as 
a  uniform  load  and,  as  is  usually  done,  the  bending  moment  for 
it  will  be  calculated  for  the  middle  of  the  bridge. 

Weight  of  i  girder  =  -*"-    -  =  6500  Ibs. 

Dead-load  bending  =  -    —  =  6500  X  45  X  —  =438,750  in.  Ibs. 
8  8 

Assuming  the  load  as  hung  centrally  on  the  trolley,  the  load  on 
each  trolley  wheel  will  be 

1  (trolley  weight  -f-  live  load)  =  J  (6000  +  20,000)  =  6500  Ibs. 
The  reaction 

_  [(6500  x  19.5)  +  (6500  x  23.5)]  =  62IO  lbs 

45 
The  maximum  bending  due  to  live  load  then  is 

M  =  6210  X  21.5  X  12  =  1,602,180  in.  Ibs. 
The  total  bending  due  to  combined  dead  and  live  loads  is 

M  —  438,750  +  1,602,180  =  2,040,930  in.  Ibs. 
174 


GIRDERS    FOR   TRAVELING   CRANES 


175 


The  material  will  be  determined  for  the  two  following  sec- 
tions, the  depth  being  taken  as  3  feet. 


H 


FIG.  205.  FIG.  206.  FIG.  207. 

The  box  section  (Fig.  207)  will  be  considered  first.  If  the 
width  is  made  one-thirtieth  the  span  it  will  be,  say,  17  ins.  wide. 
The  ends  of  the  girders  resting  upon  the  bridge  trucks  will  be 
assumed  18  ins.  deep.  The  web  plates  will  be  made  \  in.  thick, 
making  the  gross  web  area  at  the  ends  2  X  J  X  18  =  9  sq.  ins. 


FIG.  208. 


The  maximum  end  shear  is  given  by  the  diagram  of  maximum 
shears  (Fig.  208),  and  is  15,800  Ibs.     The  unit  shearing  fiber 


stress  is 


=   1760  Ibs.     This  fiber  stress  is  very  low  but, 


1  76  GRAPHICS  AND   STRUCTURAL  DESIGN 

notwithstanding  that  some  specifications  permit  i^-in.  material, 
we  will  use  the  J-in.  plates. 

Compression  Flange.  —  The  distance  between  the  centers  of 
gravity  of  the  flanges  will  be  assumed  as  approximately  the 
distance  back  to  back  of  the  flange  angles,  or  36  ins.  According 
to  the  curve,  page  100,  the  allowable  fiber  stress  in  a  compression 
flange  when  the  ratio  of  the  laterally  unsupported  length  of 
flange  to  flange  width  is  30  to  i,  is  80  per  cent  of  the  maximum 
desired.  The  allowable  fiber  stress  becomes  10,000  X  0.80  = 
8000  Ibs.  per  sq.  in. 

The  net  flange  area  then  becomes  (see  page  146), 

M 


If  the  web  is  considered  as  taking  bending  in  addition  to  shear 
the  web  will  furnish  a  section  equal  to  one-eighth  the  web  area,  or 
f  X  36  X  1  X  2  =  2.25sq.  ins.  In  this  case  the  flange  plates  will 
work  out  so  light  that  the  web  plates  will  not  be  assumed  as  re- 
sisting bending.  The  flange  angles  will  be  tried  at  3  ins.  X  3  ins. 
X  \  in.  and  they  will  be  considered  as  having  two  holes,  one  in 
each  leg  at  the  same  section;  this  makes  it  possible  to  locate  the 
rivets  at  any  points  desired.  The  plate  will  be  taken  as  17  ins. 
wide.  Net  area  of  angles,  3  ins.  X  3  ins.  X  i  in.,  having  two  f-in. 
diameter  holes, 

(2  X  1.44)  —  (4  X  0.22)  =  2.88  —  0.88  =  2.oosq.  ins. 

The  net  area  of  the  plate  then  is  7.1  —  2.00  =  5.10  sq.  ins.  The 
net  width  of  the  plate  is  17  —  (2  Xf)  =  15.25  ins.  The  thick- 

ness of  the  plate  is  —  —  =  0.334  in.  ;  use  A-  in. 

I5-2S 
Tension  Flange  Area.  —  Here   the  allowable  fiber  stress  is 

10,000  Ibs.  per  sq.  in.  and  the  required  net  area  of  the  flange  is 

M          2,040,930 

A  =  -  —  -  =  —  '       'vo   „  =  5.67  sq.  ins. 
fXh      10,000  X  36 

Using  two  3  in.  X  3  in.  X  J-in.  angles  as  before,  the  net  area  of  the 


GIRDERS  FOR  TRAVELING   CRANES  177 

plate  is  5.67  —  2.00  =  3.67  sq.  ins.  and  the  required  thickness  of 
the  plate  is  -^-^  =  0.241  in.,  say  \  in. 

The  girders  will  be  made  approximately  for  uniform  strength 
(see  Fig.  209).  The  depths  at  points  along  the  girder  may  be 
checked  by  drawing  the  diagram  of  maximum  bending  moments 
similar  to  that  drawn  for  a  railway  girder,  in  Chapter  XI,  and 
combining  the  live-load  bending  moments  thus  found  with  the 
dead-load  bending  moments. 


FIG.  209. 

Since  M  =  flange  area  X  mean  fiber  stress  X  girder  depth, 
it  follows  that  the  flange  area  and  the  mean  fiber  stress  being 
made  constant  the  depths  at  any  section  can  be  made  to  vary 
with  the  bending  moments.  This,  of  course,  is  approximately 
true  only  so  long  as  the  flange  area  times  the  square  of  its  dis- 
tance from  the  neutral  axis  of  the  entire  girder  section  at  the 
point  considered  is  large  compared  with  the  inertia  of  the  flange 
about  a  parallel  axis  through  its  center  of  gravity.  Where 
greater  accuracy  is  desired  the  moments  of  inertia  of  the  several 

sections  may  be  computed  and  the  formula  M  =  /  -  used. 

(s 

The  section  chosen  should  now  be  tested  when  subjected  to  the 
lateral  moment  stated  in  the  specification.  This  moment  being 
produced  by  one-tenth  of  the  live  load  it  will  equal  one-tenth  of 

the  live-load  moment,  or =  160,218  in.  Ibs.     The  web 

10 

plates  for  a  distance  of  12  ins.  from  the  top  will  be  considered  as 


i78 


GRAPHICS  AND   STRUCTURAL  DESIGN 


GIRDERS   FOR  TRAVELING   CRANES 


170 


resisting   this   moment.     The   following   calculation   gives   the 
moment  of  inertia  about  the  axis  B-B,  Fig.  211. 


T 


1 


5/«JPlate 


3x3x&L 
FlG.  211. 


-4-B— 


One  plate,  — 


i 26 . 90 


Two  plates,  AW-  =  2  X  0.25  X  (12  -  0.88)  X  (S-^)2  = 145-73 

Two  angles  (about  their  own  axis)  = 2 . 48 

Two  angles,  AW-  =  2  X  1.44  X  6.09*  = 106 . 82 


Deducting  for  rivet  holes  (Ah2  portion  only), 

2  X  0.22  X  S-382  = 12 . 73 

2  X  (0.27  +  0.22)  X  6.752  = 44-65 


Inertia 


Inertia 


381-93 


57.38 
324-5S 


The  extreme  fiber  stress  is  then  /  = 


M  -e 


or 


/ 


8. 


160,218  X    ~'J    =  4200  Ibs.  per  sq.  in. 
324.6 


The  mean  fiber  stress  in  the  compression  flange  is 
M         2,040,930 


/= 


=  8420  Ibs.  per  sq.  in. 


Axh      6.73X36 

The  total  fiber  stress  is  8420  +  4200  =  12,620  Ibs.  per  sq.  in. 

This  fiber  stress,  considering  that  it  is  the  maximum  of  both 
vertical  and  lateral  bending,  is  satisfactory. 

Some  designers  prefer  to  estimate  the  forces  acting  laterally 
on  the  crane  upon  the  assumption  that  the  crane  when  carrying 
its  maximum  load  is  stopped  from  full  speed  in  an  assumed  dis- 


i8o  GRAPHICS  AND   STRUCTURAL  DESIGN 

tance,  the  retardation  being  considered  uniform.     In  the  present 
instance  we  will  assume  a  velocity  of  bridge  travel  of  300  ft.  per 
minute,  and  that  the  crane  is  stopped  in  5  ft. 
The  crane  velocity  being  ^-  =  5  feet  per  second,  the  retarda- 


tion  is  /  =  --  ,  where  /  =  acceleration  or  retardation  in  feet  per 

second  per  second. 

v  =  velocity  of  bridge  in  feet  per  second. 

s  =  distance  traveled  during  acceleration  or  retardation  in 
feet. 

-2 

In  our  case  the  retardation  is/  =  —  ^—  =  2.5  ft.  per  sec.  per  sec. 
As  the  force  equals  mass  times  retardation,  the  force  required  to 

stop  the  bridge  is  F  =  —  -  X  2.5  =  1000  Ibs. 

32.2 

This  acts  as  a  uniform  load  laterally  along  the  bridge  and  will 
be  resisted  by  both  top  and  bottom  flanges  of  each  girder.  The 
force  to  retard  the  trolley  and  live  load  will  act  upon  the  upper 

flanges  of  the  two  girders  of  the  bridge  and  equals  F  =  - 

32.2 

X  2.5  =  2020  Ibs. 

The  central  load  on  one  girder  the  equivalent  of  the  above 
forces  is  ^Y-0-  +  -^  =  1260  Ibs. 

The  bending  moment  due  to  this  central  load  is 

W  'L  12 

M  =  --  =  1260  X  45  X  —  =  170,100  in.  Ibs. 
4  4 

This  moment  differs  so  slightly  from  the  moment  found  by  the 
previous  method  that  the  girder  dimensions  should  prove  satis- 
factory viewed  from  either  estimate. 

Flange  Rivets.  —  Considering  that  the  unit  fiber  stress  in 
tension  is  10,000  Ibs.  per  sq.  in.  the  allowable  unit  shearing  fiber 
stress  on  the  rivets,  assuming  about  the  same  material,  would 
be  |  X  10,000  =  7500  Ibs.  per  sq.  in.  The  bearing  value  would 
be  taken  at  twice  the  shearing  value  or  15,000  Ibs.  per  sq.  in. 


GIRDERS   FOR   TRAVELING   CRANES  l8l 

The  value  of  f-in.  rivets  in  single  shear  and  bearing  in  a  J-in. 
plate  is  2813  Ibs. 

Assuming  the  girder  depth  of  the  supports  at  18  ins.,  the  dis- 
tance center  to  center  of  rivets  vertically  is  18  —  (2  X  1.75) 
=  14.5  ins. 

The  maximum  end  shear  will  be  assumed  as  transferred  to  the 
girder  in  a  distance  equal  to  its  depth,  or  18  ins.  The  vertical 

i  ^  800 

shear  per  inch  of  span  is  Si  =  - =  880  Ibs. 

18 

The  horizontal  shear  per  inch  at  this  point  is 

Vertical  shear 

Distance  center  to  center  of  rivets  vertically 

15.800 
$2  =  -iL-  ~  =  1090  Ibs. 

14.5 

The  resulting  shear  per  inch  of  span  is 


sz  =      Si2  -f  522  :  =  1400  Ibs. 

Considering  that  the  web  being  double  will  necessitate  a  row  of 
rivets  in  each  web  plate  the  horizontal  distance  center  to  center 

,     .  2813   X   2 

of  rivets  = =  4  ins. 

1400 

Owing  to  the  concentration  of  loading  at  the  ends  the  spacing 
will  be  made  3^  ins. 

The  net  area  of  the  compression  flange  being  6.73  sq.  ins.  and 
the  mean  fiber  stress  8750  Ibs.  per  sq.  in.,  the  number  of  rivets  to 

develop  the  full  flange  strength  is6'73  X  875°,  about  21. 

2813 

These  rivets  placed  in  two  rows  and  spaced  3^  ins.  center  to 
center  would  require  a  distance  of  3.5  X  n  =  38.5  ins.  This 
would  indicate  that  for  about  48  ins.  from  the  ends  of  the  plate 
the  rivets  should  be  spaced  about  3!  ins.  center  to  center  and  at 
the  other  sections  the  spacing  in  the  compression  flange  should 
not  exceed  from  16  to  20  times  the  thickness  of  the  outside  plate. 
The  spacing  in  the  two  flanges  is  usually  made  the  same.  The 


182  GRAPHICS  AND  STRUCTURAL  DESIGN 

upper  plate  being  T5g  in.  thick  the  rivet  spacing  should  not 
exceed  5  ins. 
Angle  Stiffeners.  — Allowing  10,000  Ibs.  per  sq.  in.  the  area  of 

..„  Maximum  end  shear        1^,800 

the  stiffeners  =  —  — •  =  1.58  sq.  ms. 

10,000  10,000 

Using  two  angles  the  smallest  permitted  would  be  i\  ins.  X 
i\  ins.  X  J  in.,  and  their  section  2  X  1.19  =  2.38  sq.  ins.,  which 
is  more  than  ample. 

Stiffeners,  however,  serve  a  double  purpose,  as  the  opposite 
inside  Stiffeners  are  riveted  to  a  plate  forming  a  diaphragm  at 
the  section  and  greatly  stiffening  the  girder  laterally.  They 
frequently  also  serve  to  secure  short  channel  or  angle  sections 
which  pass  across  the  flange  and  assist  the  flange  plate  in  carry- 
ing the  rail.  These  latter  are  commonly  placed  at  each  pair  of 
vertical  Stiffeners,  while  the  diaphragms  are  generally  placed  at 
alternate  panels. 

The  web  plate  should  be  reinforced  where  brackets  or  motors 
are  attached  to  it  and  at  these  points  handholes  should  be  cut 
in  the  web  plate  on  the  opposite  side  of  the  girder  to  facilitate 
bolting  the  brackets  and  motor  to  the  girder. 

GIRDER  DESIGN  FOR  FIG.  206 

Using  the  same  data  the  girder  will  now  be  designed  for  the 
section  (Fig.  206). 

The  maximum  combined  live-  and  dead-load  bending  was 
found  to  be  2,040,930  in.  Ibs.  The  maximum  end  reaction  pre- 
viously determined  was  15,800  Ibs.  The  girder  will  be  assumed 
36  ins.  deep,  back  to  back  of  angles,  at  the  center  of  the  span,  and 
20  ins.  deep  at  the  bridge  trucks.  The  web  will  be  taken  -f^  in. 

thick  making  the  unit  shearing  fiber  stress  *•*'         X  20  =  2550 

0.31 

Ibs.  per  sq.  in. 

The  effective  depth  will  be  assumed  slightly  less  than  36  ins. ' 
to  allow  for  the  centers  of  gravity  of  the  channels  falling  towards 


GIRDERS   FOR   TRAVELING    CRANES  183 

the  center  of  the  girder,  the  distance  being  36  —  1.5  =  34.5  ins. 
The  net  flange  area  intension  then  will  be 

M  2,040,030 

A  =  -  -   =  5.02  sq.  ins. 

fXh      10,000  X  34-5 

Trying  one  g-in.  channel  at  13!  Ibs.  per  ft.,  its  area  is  3.89  sq. 
ins.,  and  two  3  in.  X  3  in.  X  i^-in.  angles,  we  have 

Sq.  ins.  Sq.  ins. 

One  g-in.  channel  at  13!  Ibs.,  gross  area  .........     3.89 

Less  two  rivet  holes  in  £-in.  material,  2  X  .22  =  .  .        .44  3.45 

Two  3  X  3  X  iVin.  angles,  1.78  X2=  ..........     3.56 

Less  four  rivet  holes  in  tV"1-  plate,  4  X  .27  =  ...      1.08  2.48 

5-93 

This  net  area  is  satisfactory. 

In  the  compression  flange  the  fiber  stress  must  be  reduced  to 
provide  for  lateral  strength.  The  ratio  of  span  to  flange  width 
if  a  i5-in.  channel  is  tried  is  45  X  yf  =  36.  Reducing  the  unit 
working  fiber  stress  to  75  per  cent  of  the  maximum  desired 
permits  a  fiber  stress  of  7500  Ibs.  per  sq.  in.  Again  substituting 
in  the  formula 

=  7.88  sq.  ins. 


fXh      7500  X  34-5 

Trying  two  3  in.  X  3  in.  X  iVm-  angles  and  one  i5-in.  channel  at 
33  Ibs.  per  ft., 

Sq.  ins.  Sq.  ins. 

One  i5-in.  channel  at  33  Ibs  ..................     9  .  90 

Less  two  rivet  holes,  2  X  .88  X  .40  =  ..........  70  9  .  20 

Two  3  X  3  X  rVin-  angles,  1.78  X  2  =  ........     3.56 

Less  four  rivet  holes  ......  ..................     i  .  08  2  .  48 

Total  area  .............  n.68 

This  seems  to  be  too  large  but  before  discarding  it  the  combined 
stresses  due  to  the  lateral  and  vertical  bending  should  be  deter- 
mined. The  flexural  fiber  stress  with  this  flange  is 

M  2.040, 

11.68  X 


1 84  GRAPHICS  AND   STRUCTURAL  DESIGN 

In  the  preceding  girder  section  the  lateral  bending  due  to  stop- 
ping the  crane  in  5  ft.  was  estimated  at  170,100  in.  Ibs.  The 
fiber  stress,  assuming  that  the  bending  is  mainly  resisted  by  the 
i5-in.  channel  whose  section  modulus  about  its  axis  i-i  is  41.7,  is 

f      M*e      170,100 

/  =  — —  =  — — =  4100  Ibs.  per  sq.  in. 

/  41.7 

The  combined  maximum  stress  then  is 

/5ioo\   . 

(       —  )  +  4100  =  10,900  Ibs.  per  sq.  m. 

This  is  low  considering  that  everything  of  importance  is  taken 
into  account,  but  will  be  used,  as  calculations  made  for  i2-in. 
channels  indicate  a  fiber  stress  undesirably  high.  This  section 
of  the  girder  may  now  be  checked  by  calculating  its  moment 

of  inertia  and  using  the  formula  /  =  —  • 

Flange  Riveting.  —  These  calculations  are  similar  to  those 
made  for  the  box  girder.  The  rivet  value  in  a  yVm-  web  plate 
for  f-in.  rivets  is  3516  Ibs.  As  in  the  other  case 

15,800 

51  =  -^-  ~  =  790  Ibs. 

20 

15,800   ,  ,, 

52  =  -~ =  960  Ibs. 

16.5 

The  resulting  shear  per  inch  of  span  is 


53  =  vV  +  s22  =  V79o2  +.  96o2  =  1240  Ibs. 

and  the  rivet  spacing  of  f-in.  rivets  is  ffrl  =  2.83  ins. 

The  stiffeners  can,  as  in  the  preceding  problem,  be  made  the 
smallest  allowable,  say,  2\  in.  X  2|  in.  X  |-in.  angles. 

Bridge  Girders  with  Horizontal  Stiffening  Girders.  —  The 
following  example  is  intended  to  illustrate  this  type  of  bridge. 
Span  60  ft.  Girder  depth  f|-  =  5  ft.  back  to  back  of  angles. 
The  crane  capacity  is  to  be  20  tons  (40,000  Ibs.).  The  trolley 


GIRDERS   FOR   TRAVELING  CRANES  185 

wheel  base  is  6  ft.  o  ins.     The  bridge  weight,  two  girders,  is 
30,000  Ibs.     The  load  on  each  trolley  wheel  is 

4O.OOO  +  I2.OOO 

-  =  13,000  Ibs. 
4 

The  live-load  bending  will  be  found  as  in  the  preceding  cases 
and  as  follows: 

Rl  .  (13,000  X  3I..S)  +  (13.000  X  25.5)  .  lbg 

oo 

M  =  R!  X  28.5  X  12  =  12,350  X  28.5  X  12, 
M  =  4,223,700  in.  Ibs. 

k^U< 25.5^ 

4$- 


FIG.  212. 

In  estimating  the  dead-load  bending  the  girder  weight  will  be 
assumed  as  a  uniform  load.  The  bending  moment  is 

,,      WL      15,000  X  60  X  12 
M  =    ——  ——          -  =  1,350,000  in.  Ibs. 

o  o 

The  total  bending  is  4,223,700  +  1,350,000  =  5,573,700^1.  Ibs. 

Allowing  a  mean  unit  fiber  stress  of  9000  Ibs.  per  sq.  in.  the 
flange  area  will  be 

M        <s,  =571,  700 

A  =7~L=  '      o    =  IO-65  sq.  ms. 

/  •  h      9000  X  58 

The  distance  between  the  centers  of  gravity  of  the  flanges  has 

been  taken  58  ins.  as  probably  no  flange  plates  will  be  required. 

The  web  plate  will  be  assumed  as  J  in.  thick  and  one-eighth  the 

web  area  will  be  considered  as  contributing  to  the  flange  section. 

Web  area  =  60  X  i  =  15  sq.  ins. 

|  X  web  area  =  ^  =  1.88  sq.  ins.  The  balance  of  the  flange 
area  will  be  made  up  of  the  angles  or,  10.65  ~~  J-88  =  8.77  sq. 
ins. 


i86 


GRAPHICS   AND   STRUCTURAL  DESIGN 


The  flange  having  two  angles  each  angle  must  have  a  net  area 


of  -       =  4.38  sq.  ins.     Allowing  for  two  holes  for  J-in.  diameter 

rivets,  the  gross  area  of  each  angle  must  be  4.38  +  0.41  =  4.79 
sq.  ins. 

Trying  6  in.  X  4  in.  X  J-in.  angles,  their  gross  area  is  4.75 
sq.  ins.  which  is  satisfactory.  The  flange  will  therefore  be  made 
of  two  6  in.  X  4  in.  X  J-in.  angles. 

Horizontal  Stiffening  Girder  (Fig.  213).  —  First  assuming  a 
horizontal  loading  of  one-tenth  the  vertical  loading  gives  two 

FIG.  213. 


concentrated  loads  of  1300  Ibs.  each  and  a  uniform  load  of  1500 
Ibs.  Before  using  these  figures  we  will  check  them  by  estimating 
the  forces  acting  on  the  girders  when  the  crane  carrying  its  full 
load  at  maximum  velocity  is  stopped  within  5  ft. 

The  velocity  of  the  crane  being  assumed  at  300  ft.  per  minute, 
or  5  ft.  per  second,  to  stop  the  crane  in  5  ft.  the  retardation  must 

be  /  =  -   -  =  —  X  5  =  2.5  ft.  per  sec.  per  sec.     The  force  to 

2  •  S  2 

check  the  live  load  and  trolley  is  Force  =  Mass  X  Acceleration. 

,-,       52,000  X  2.5 
F  =  —  — -  =  4040  Ibs. 

32.2 

The  force  to  check  the  bridge  is 

p  =  30,000  X  2.5  Ib5. 

32.2 


GIRDERS  FOR  TRAVELING   CRANES 


187 


The  first  -^-054—  =  1010  Ibs.  compares  with  a  concentrated  load  of 
1300  Ibs.,  while  the  latter  is  a  uniform  load  of  1165  Ibs.  compared 
with  one  of  1 500  Ibs.  The  first  figures,  being  the  greater,  will  be 
used. 

As  this  lateral  bending  is  at  best  only  a  rough  approxima- 
tion it  will  be  sufficiently  accurate  to  assume  a  central  load  of 
2  X  1300  =  2600  Ibs.,  due  to  trolley  weight  and  live  load  and 
an  equivalent  central  load  of  -52°—  =  750  Ibs.,  due  to  bridge 

weight.     The  reaction  then  isf-  '5  j  =  1675  Ibs. 

The  stress  AT,  Fig.  214,  is  10,300  Ibs. 

A  section  through  the  main  and  bracing  girders  is  shown  in 
Fig.  215. 


FIG.  216. 


As  the  wheels  travel  along  the  flange  of  the  main  girder  the 
force  of  1300  Ibs.  transferred  to  the  girder  by  a  wheel  may  pro- 
duce bending  on  this  flange,  for  instance,  between  the  two  points 
of  lateral  support  i  and  2  this  bending  will  be 

W-L  60 

M  =  -      -  =  1300  X  —  =  19,500  in.  Ibs. 

4  4 

The  moment  of  inertia  of  two  6  in.  X  4  in.  X  J-in.  angles  back 
to  back  but  separated  by  a  J-in.  plate  is 

Angles  about  their  own  axes,  2  X  17.40  = 34 .80 

A. W  =  2  X  4-75  X  (i.Q9  +  0.12)2  = 42.27 


/  =    77.07 
The  fiber  stress  due  to  bending  is 

f  •  e      iQ,S°°  X  6.12 

-  =    ^  °  -  =  1550  Ibs.  per  sq.  in. 

I  77.07 


/  _ 


1  88  GRAPHICS   AND   STRUCTURAL  DESIGN 

Direct  fiber  stress  due  to  horizontal  bracing  girder 
,  10,300 

5 


(2  x  4.75) 

The  total  fiber  stress  then  is 

9000  ;+-  1085  +  1550  =  11,635  Ibs.  per  sq.  in. 

As  all  the  forces  acting  on  the  girder  both  vertically  and  hori- 
zontally have  been  considered  this  fiber  stress  is  satisfactory. 
The  unbraced  length  of  A  I  being  60  ins.  the  radius  of  gyration  if 

-  =  120  then  r  =  —  •  =  -•     The  smallest  angles  fulfilling  these 
r  120      2 

requirements  are  3  ins.  X  3  ins.     Trying  one  3.5  in.  X  3.5  in.  X  yV 

in.  angle,  its  least  radius  of  gyration  is  0.69.      -  =  -   -  =  87. 

r      0.69 

According  to  Ritter's  formula, 

r  11,000 


10,000 

The  total  load  on  a  3.5  in.  X  3.5  in.  X  yVm-  angle  is  6250  X  2.09 
=  13,000  Ibs.     This  satisfies  the  requirements. 

The  diagonals  will  carry  their  maximum  stresses  when  the 
trolley  fully  loaded  is  at  one  side  of  the  bridge.  The  maximum 
end  shear  on  the  horizontal  girder  is 

„   2600  X  S7  , 

Ri  =  -      -  +  750  =  2470  H-  750  =  3220  Ibs. 

DO 

The  force  in  the  diagonals  then  is  3220  X  1.41  =  4550  Ibs. 
The  length  of  the  diagonal  is  60  X  1.41  =  84.6  ins.  If  -  =120 

then  r  =  -    —  -0.70.     This  will  require  3^  in.  X  3f-in.  angles. 
1  20 

The  permissible  fiber  stress,  according  to  Ritter's  formula,  is 

f  II  ,OOO  1  1  ,OOO 

/  =  -  ~  -  Tjr,  =  -      r  -  4Soo  Ibs.  per  sq.  in. 
i  +-       -  X(-|       i  + 


10,000      \rj  10,000 


GIRDERS  FOR   TRAVELING   CRANES 


189 


The  load  that  can  be  carried  by  one  3^  in.  X  3^  in.  X  yVm-  angle 
is  4500  X  2.09  =  9400  Ibs. 

No  calculation  will  be  made  for  the  vertical  latticed  girder  as 
it  merely  holds  the  horizontal  girder  in  place;  the  smallest  per- 
missible angles,  2\  ins.  X  2^  ins.  X  \  in.  will  be  used.  The  lower 
horizontal  girder  will  be  made  the  same  as  the  upper,  excepting 
that  the  angles  at  right  angles  to  the  main  girder  will  be  omitted. 

Riveting.  —  To  determine  the  rivet  spacing  in  the  main  gir- 
der it  will  be  necessary  to  estimate  the  maximum  shears  along 


r 


FIG.  217. 

the  girder.     The  maximum  reactions  on  one  girder  are,  live- 
load  reaction,  26,000  X  f  -J  =  24,700  Ibs.     The  dead-load  reaction 


is 


=  7500  Ibs.     The  combined  live-load  and  dead-load 


reactions  are  24,700  +  7500  =  32,200  Ibs.  See  Fig.  217,  which 
is  the  diagram  of  maximum  shears  for  this  girder.  The  maxi- 
mum wheel  load  will  be  assumed  as  transferred  to  the  web  in 
24  ins.  of  flange  length,  or  the  shear  per  inch  of  span  due  to  load 


concentration  is 


24 


=  540  Ibs.  =  Si. 


190  GRAPHICS   AND   STRUCTURAL   DESIGN 

The  horizontal  shear  at  the  ends  per  inch  of  span  is, 

End  shear  divided  by  the  vertical  distance  center  to  center  of 

n  •  32,200  32,200 

flange  rivets  = ^ rr  =  -       -  =  585  lbs.  =  s2. 

[60  —  (2  X  2.5)]          55 

53  =  vV  +  s22  =  V54o2  +  5852  =  796  lbs. 

The  rivet  value  of  a  f -in.  rivet  in  double  shear,  bearing  in  a 
J-in.  plate,  is  2813  lbs.  for  a  unit  shearing  stress  of  7500  lbs.  per 
sq.  in.  and  15,000  lbs.  per  sq.  in.  in  bearing.  The  rivet  spacing 
therefore  will  be  -^/-  ~  3!  ins. 

The  rivet  spacing  will  be  determined  for  quarter  points  along 
the  girder,  see  Fig.  218.  At  15  ft.  from  the  end  the  maximum 


shear  is  22,000  Ibs.  and  the  horizontal  shear  per  inch  of  span  is 

22,000  „ 

-  =  400  Ibs. 

s3  =  \/4oo2  +  5402  =  672  Ibs. 

The  rivet  spacing  is  -2gV2A  =  4-I8  ins. 

At  the  middle  the  shear  is  11,700  Ibs.  and  the  horizontal  shear 

.   11,700 
per  inch  of  span  is  -     —  =  213  Ibs.,  hence 


=  580  Ibs. 

The  rivet  spacing  is  -Vro3-  =  4.85  ins. 

From  the  diagram,  Fig.  218,  the  rivet  spacing  for  any  point 
along  the  girder  is  readily  scaled, 


CHAPTER  XIV 
REINFORCED   CONCRETE 

REINFORCED  concrete  is  designed  by  an  extension  of  the 
principles  applied  to  the  discussion  of  structural  steel.  The  modi- 
fication is  necessary  since  reinforced  concrete  is  a  composite  of 
several  materials  and  therefore,  unlike  steel,  is  nonhomogeneous. 

In  ordinary  practice  the  concrete  varies  from  1,2,4(1  cement, 
2  sand  and  4  stone,  slag  or  gravel)  to  i,  3,  6.  The  best  mixture 
depends  upon  the  character  of  the  materials,  the  purpose  being 
to  have  the  cement  fill  the  voids  in  the  sand,  while  the  mortar 
thus  produced  shall  fill  the  voids  in  the  stone,  slag  or  gravel. 

Cement.  —  Portland  cement  only  should  be  used.  This  is 
practically  the  only  cement  available  as  very  little  natural 
cement  is  now  manufactured. 

The  cement  should  meet  the  usual  standard  specifications. 

Sand.  —  The  sand  should  be  a  clean,  coarse,  sharp  sand  free 
of  clay  and  dirt.  Fine,  rounded  or  dirty  sand  greatly  reduces 
the  strength  of  the  concrete. 

Stone  and  Gravel.  —  Stone  is  usually  broken  so  that  the 
maximum  size  does  not  exceed  from  f  in.  to  ij  ins.  Stone  and 
gravel  should  be  freed  from  dust,  sand  and  dirt,  but  uniformity 
of  size  is  not  desirable  as  it  increases  the  percentage  of  voids. 

Mixing.  —  Practice  formerly  demanded  a  fairly  dry  mixture, 
tamped  until  the  water  came  to  the  top,  but  present  practice 
uses  a  very  wet  mixture,  which  greatly  reduces  the  tamping 
required  and  makes  a  more  solid  concrete.  The  wet  concrete 
is  weaker  than  the  other  when  new,  but  its  strength  rapidly 
approaches  that  of  the  other  with  age. 

Physical  Properties.  —  When  one  month  old  and  made  from 
ordinary  materials  the  compressive  strength  of  i,  2,  4  stone 

concrete  can  be  assumed  at  from  1800  Ibs.  to  2200  Ibs.  per  sq.  in., 

191 


192 


GRAPHICS  AND   STRUCTURAL  DESIGN 


while  a  i,  3,  6  mixture  will  have  a  strength  of  about  20  per  cent 
under  these  figures. 

The  modulus  of  elasticity  increases  with  age  and  decreases 
with  the  unit  stress,  so  that  only  approximate  values  can  be 
assumed.  A  commonly  accepted  modulus  of  elasticity  for  mak- 
ing reinforced-concrete  calculations  is  2,000,000  Ibs.  per  sq.  in., 
being  one-fifteenth  that  of  steel. 

Elastic  Limit.  —  There  is  no  true  elastic  limit  for  concrete 
as  it  shows  permanent  deformation  under  the  lightest  loads. 
According  to  Bach  and  others  it  seems  fair  to  assume  that 
what  is  generally  accepted  as  the  elastic  limit  in  concrete  occurs 
at  a  unit  stress  of  from  one-half  to  two-thirds  of  the  ultimate 
compressive  strength. 

PHYSICAL  PROPERTIES  OP  STONE  OR  GRAVEL  CONCRETE 


Age,  Days. 

i,  2,  4  Mixture. 

I,  3,  6  Mixture. 

Compressive  strength.  .  .  . 
Tensile  strength  

30 
3O 

I,8oo-2,2OO 
150—200 

1,450-1,750 
100-125 

Shearing  strength  

3° 

1,000—1,400 

9OO—I,IOO 

Modulus  of  elasticity  
Coefficient  of  expansion, 
i°F. 

90 

2,500,000-3,500,000 
o  0000060 

2,500,000-3,500,000 
o  0000060 

Weight  per  cubic  foot  in- 
cluding steel  

150 

ISO 

NOTE.  —  All  stresses  in  pounds  per  square  inch.  Slag  or  cinder  concrete  will  only 
develop  about  one-third  the  strength  of  a  good  stone  concrete.  Its  modulus  of 
elasticity  will  range  from  1,000,000  for  lean  mixtures  up  to  2,500,000  for  i,  2,  4 
and  richer  mixtures.  It  weighs  about  120  Ibs.  per  cu.  ft. 


USUAL  UNIT  WORKING  STRESSES.    FLEXURE 

Extreme  fiber  stress  on  concrete,  compression 

Extreme  fiber  stress  on  concrete,  tension 

Steel,  soft,  tension 

Steel,  mild  and  hard,  tension 

Bonding  stresses,  straight  bars 

Bonding  stresses,  straight  bars,  over  supports 

Bonding  stresses,  bent  or  twisted  bars 

Shearing,  plain  web,  no  steel  stirrups 

Shearing,  when  reinforced  with  stirrups,  or  bars 

Tension,  diagonal,  on  concrete 

Modulus  of  elasticity  of  concrete 

Modulus  of  elasticity  of  steel 


Lbs.  per  sq.  in. 
500-650 
O 

12,000 
16,000 

60-80 

90-120 

150 

30 

100 

30 

2,000,000 
30,000,000 


REINFORCED   CONCRETE  193 

USUAL  UNIT  WORKING  STRESSES.     COLUMNS 

Columns  not  reinforced 450 

Columns,  reinforced  longitudinally  only 450 

Columns,  reinforced  with  hoops  or  bands 550 

Columns,  reinforced  with  hoops  or  bands,  and  with  from  i 

to  4  per  cent  of  longitudinal  steel 650 

Columns,  structural  steel  thoroughly  inclosing  concrete 650 

In  general  the  following  nomenclature  and  formulae  conform  to 
the  report  of  the  Joint  Committee  on  Reinforced  Concrete  of 
the  American  Society  of  Civil  Engineers,  American  Society  of 
Testing  Materials,  American  Railway  Engineering  and  Main- 
tenance of  Way  Association  and  the  American  Portland  Cement 
Manufacturers. 

/,  =  tensile  unit  stress  in  steel,  pounds  per  square  inch. 

fc  =  compressive  unit  stress  in  concrete,  pounds  per  square 
inch. 

Ea  =  modulus  of  elasticity  of  steel,  pounds  per  square  inch. 

Ec  =  modulus  of  elasticity  of  concrete,  pounds  per  square  inch. 

*-* 

M  =  moment  of  resistance  or  a  bending  moment,  inch  pounds. 
A  =  area  of  steel,  square  inches. 

b  =  width  of  beam,  inches. 

d  =  depth  of  beam  to  center  of  reinforcement,  inches. 

k  =  ratio  of  depth  of  neutral  axis  to  effective  depth,  d. 

x  =  depth  of  resultant  compression  from  top. 

j  =  ratio  of  lever  arm  of  resisting  couple  to  depth,  d. 
j  •  d  =  d  —  x  =  arm  of  resisting  couple. 

p  =  ratio    of    steel    to    concrete.      For    rectangular   beams, 


^4 
B  =  width  of  flange  of  T  beams,  inches.     Here  p  — 


B-d 

b1  .=  width  of  stern  of  T  beams,  inches. 
t  =  thickness  of  flange  of  T  beams,  inches. 
_/.  XEC 
~fcXE.' 


194  GRAPHICS  AND  STRUCTURAL  DESIGN 

SHEAR  AND  BOND  STRESSES 
V  —  total  shear,  pounds. 

v  =  shearing  unit  stress,  pounds  per  square  inch. 
u  =  bond  stress,  pounds  per  square  inch  of  bond  area. 
o  =  circumference  or  perimeter  of  bar,  inches. 
S0  =  sum  of  the  perimeters  of  all  bars,  inches. 

BEAMS  WITH  DOUBLE  REINFORCEMENT 
A1  =  area  of  compression  steel. 
pl-=  steel  ratio  of  compression  steel,  not  percentage. 
fs1  =  unit  compression  in  steel. 

C  =  total  compression  in  concrete. 
C1  =  total  compression  in  steel. 

dl  =  depth  to  center  of  compressive  steel. 

e  =  depth  to  resultant  of  C  and  C1. 

»-  j 

COLUMNS 

A  =  total  net  area. 
A  8  =  area  of  longitudinal  steel. 
Ac  =  area  of  concrete. 

P  =  total  safe  load. 

THEORETICAL  DISCUSSION  or  REINFORCED-CONCRETE  BEAMS 
Rectangular  Beams.  —  The  usual  formulae  for  the  design  of 
reinforced-concrete  beams  are  developed  upon  the  assumptions, 
(a)  that  a  cross-section  plane  before  bending  continues  a  plane 
during  bending,  and  (b)  that  stress  and  strain  are  proportional. 

From  these  assumptions  the  strain  in  compression  a  unit 
distance  above  the  neutral  axis  must  equal  that  in  tension  a  unit 
distance  below  that  axis,  or,  from  Fig.  219, 

fc  fs  _  fs  _  fs 

kdxEc      (d-kd)E8      (jd-%kd)E9     d(j-lk) 
fsXEc  Jd-lkd  =  d-kd  = 
fc  X  E,  kd  kd 


from  which 


and 


REINFORCED   CONCRETE 
*---i- 


y-^ 


3  («  + 


(i) 
(2) 


According  to  the  principles  of  mechanics  the  sum  of  the  hori- 
zontal forces  acting  on  a  vertical  section  must  be  zero,  hence  the 
resultant  force  in  the  compression  flange  must  equal  the  result- 
ant force  in  the  tension  flange,  or 


(3) 


and  the  moments  are  also  equal,  M8  =  Jlf  c. 
Ma  =  A  Xf.Xj-d. 


MC=J- 

2 


-"xk.jxb.f.         (4) 

2 


a  b 


kd 


6      a 
FIG.  219. 

These  equations  apply  when  the  maximum  fiber  stresses  fe 
and  /,  are  known;  the  reinforcement  then  is  the  critical  ratio 
for  these  fiber  stresses. 

It  is  sometimes  desirable  to  discuss  the  beam  when  a  ratio  of 
reinforcement  is  assumed  which  is  not  necessarily  the  critical 
ratio. 

From  the  fundamental  equations, 

/  Z7  /•/•     ( -r  li\ 

k  -  d  _  t\  _£• n(i-fc) 

f,~(l      k)k.Ec  k 

From  equations  (3)  and  (4), 


(5) 


196  GRAPHICS  AND   STRUCTURAL  DESIGN 

n  (i  -  k)       k  k* 

hence  -A— — L  =  —      and    p  =  - 

k  2  p  2  •  n  (i  —  & 

Solving  for  &  we  find 


k  =  V 2  .  /> .  n  +  0  •  n)2  -  p  •  w.  (6) 

& 

,_,_.. 

The  fiber  stresses  may  be  found  by  substituting  these  values 
of  k  andy  in  equations  (3)  and  (4). 

APPROXIMATE  FORMULAE 

The  curves  on  page  197  give  the  calculated  relations  between 
z,  k,  j,  and  n  •  p.  It  will  be  noted  that  for  the  usual  values  of  p,j 
approximates  f  and  k  approximates  f .  The  following  approxi- 
mate formulae,  using  these  values  of  k  andy,  are  sometimes  used. 

d.  (7) 

*.  (8) 

The  curves,  Fig.  220,  give  the  relations  between  the  several 
quantities. 
These  curves  are  plotted  upon  a  base  line  giving  the  values  of  z; 

therefore  when  z  =  y =r  is  known,  the  other  values  k,  j,  j  •  k, 

fc  X  h>s 

p  •  n  and  p  are  read  on  the  perpendicular  to  z.  Thus  f  or  z  =  2.2 
the  values  are  read  on  the  dotted  line  a-a  through  2.2.  The 
values  are  j  =  0.895;  ^  =  °-3°7>  ^  "j  =  °-275J  P  *n  =  0.072 
and  p  for  n  =  15  is  p  =  0.0048. 

Had  the  value  of  p  i  or  n  =  15  been  given  as  0.0048  instead  of 
the  value  of  z  =  2.2  the  horizontal  line  bb  would  have  been 
drawn  through  p  =  0.0048  until  it  cut  the  curve  of  p  for  n  =  15, 
and  through  this  point  of  intersection  the  line  aa  would  be 
drawn  perpendicular  to  the  z  axis  and  upon  this  line  aa  the 
other  values  would  be  read  as  before. 

The  following  example  will  illustrate  the  use  of  the  formulae 
and  table. 

Example. — A  rectangular  beam  in  which  d  =  22  ins.  is  to 


REINFORCED    CONCRETE 


197 


198  GRAPHICS  AND   STRUCTURAL  DESIGN 

resist  a  bending  moment  of  384,000  in.  Ibs.  Assume  n  =  15, 
/,  =  16,000  and/c  =  650;  find  the  width  of  the  beam  and  the  area 
of  the  steel. 

/.  X  Ec          16,000 


"     fcXE8      (65oXi5)  = 

Interpolating  from  the  curves,  k  =  0.379,  j  =  0.874,  k  •/ 
=  0.331  and  p  =  0.0077.  Substituting  these  values  in  equation 
(4)  gives 

,    __  2  Afc  _  2   X  384,000          _ 

~fcXk-jXd*~  650  X  0.331   X  222  ~  7'37  ^ 

A  =  bXdXp  =  7.37  X  22  X  0.0077  =  1.25  sq.  ins. 

The  solution  of  this  problem  by  the  approximate  method  gives 
almost  identical  results. 

If-   JX^i   •/.-<*, 

.  8»  M  8  X  384,000 

therefore  A  =  —  :  —  :  =  -  -  =  1.25  sq.  ins. 

7-fs-d      7X16,000X22 


M-J..t.>    or    »- 

,       6  X  384,000 
b  =  --  s2-       -  =  7.94  ins. 
600  X  222 

Another  example  will  illustrate  the  procedure  when  it  is  re- 
quired to  check  an  existing  beam.  A  reinforced-concrete  beam 
has  the  following  dimensions:  b  =  9  ins.,  d  =  20  ins.,  A  =1.50 
sq.  ins.,  n  =  15,  and  it  is  desired  that  /.  should  not  exceed 
16,000  Ibs.  per  sq.  in.  nor/c  exceed  650  Ibs.  per  sq.  in. 

p  =  A,  =  ^5^  =  O.oo825. 
b-d      (9  X  20) 

From  equation  (5), 

L  =  A  ,=  _  Q-390      _  6 

fe      2p      (2  X  0.00825) 

and    /,  =  23.6  X  650  =  15,300  Ibs.  per  sq.  in. 

This  fiber  stress  being  under  the  16,000  Ibs.  per  sq.  in.  is  satis- 


REINFORCED    CONCRETE 


199 


factory.      The   bending  moment  corresponding   to   these   fiber 
stresses  can  be  found  from  either  equation  (3)  or  (4). 
M8  =  A  Xf,  Xj  •  d  =  1.5  X  i5,3°°X  0.870  X  20  =  399,000  in.  Ibs. 
Checking  this  by  the  other  formula  gives 

Mc  =  ±(fcXk.jXb-  d2)  =  396,000  in.  Ibs. 
This  is  about  as  close  agreement  as  can  be  expected  considering 
the  curve  readings,  the  difference  being  under  i  per  cent. 

T  BEAMS 

In  reinforced-concrete  design  the  section  most  frequently  met 
is  the  T  beam,  shown  in  Fig.  221.     The  floor  is  usually  a  slab  of 


/ 

r 

r-  — 

JJi 

r 

::: 

\ 

\ 

} 

\  - 

1  1 

4Y 
I 

c 

1  i 

i  i 

i 

r 

T-r1 

1 
1 

1 
1  1 

1 

H 

N 

k  1 

j    ' 

j 

L_i 

_ 

L  

i 

" 

r 

m 

4 

M 

Colum 

. 

" 

1 

r 

r 

rr 

1 

_j 

1 

f  ^ 

M 

\ 

SG 

L  ^ 

r  r 

L  , 

TT 

/ 

[T-Beam 

( 

y 

1 

L 

l~ 

y 

L 

r 

L 

r~ 

y 

*\ 

•-T-Girc 

lc 

r 

FIG.  221. 

reinforced  concrete  laid  on  floor  beams,  corresponding  in  the 
ordinary  timber  construction  to  joists.  In  the  case  of  concrete 
construction  the  floor  beams  and  the  slab  over  it  become  one 
piece,  the  horizontal  shear  at  the  junction  of  the  two  being  re- 
sisted by  the  concrete  and  usually  by  reinforcing  stirrups  or  rods. 


2OO 


GRAPHICS   AND   STRUCTURAL  DESIGN 


When  the  neutral  axis  i-i,  in  Fig.  222,  falls  in  the  flange  or  slab 
the  formulae  and  table  previously  derived  for  rectangular  beams 

i 


FIG.  222. 


are  sufficient.     When,  however,  the  neutral  axis  falls  in  the  stem, 
as  is  usually  the  case,  another  set  of  formulae  are  required. 

GENEEAL  FORMULAE  FOR  T  BEAMS  REINFORCED  FOR  TENSION 

ONLY 

Neglecting  the  compression  in  the  stem,  and  on  the  same 
conditions  a  and  b  that  applied  to  rectangular  beams  we  have  as 
before: 


Jc  X>  -E-'a                     J  c 

Also  from  Fig.  223             j  = 

/i 

U  B  H 
1  i                     • 

k 
k-d 

•>    & 

IIU       K   - 

i 

-j 

k-d- 

t 

t 

A. 
»    •   • 

J 

"T 

X 

-El    -\ 

P 

I    1 

-i-^3 

fi     J    ; 

FIG.  223. 

Assuming  that  the  fiber  stress  in  the  concrete  varies  as  its  dis- 
tance from  the  neutral  axis,  the  distance  x  from  the  top  of  the 
beam  to  the  resultant  compression  R  is 

_/e+2/i          t          3-k-d-  2-t          tm 
•*  ~  f      s\  j       j  XN       j 

fe+fi       3         2-k-d-t        3 


REINFORCED   CONCRETE  2OI 

~ 


this  gives       3  *d  =  d  ~  x  =  d  ~  \2  .k  .d~-t 
The  average  fiber  stress  in  the  concrete  is 


It  follows  that 

Mc  =  BXtXfc(i-^j.d,  (10) 

and  Mt  =  Axf,Xj-d.  (n) 

Equation  (n)  can  be  approximated  as 


When  the  ratio  of  the  reinforcement  is  known,  not  the  fiber 
stresses,  equation  (i)  cannot  be  used  to  find  k  as  z  is  not  known. 
Taking  equations  (10)  and  (n),  we  find, 

fc=_A_      i_  pd  k 

I~7kd      \~2~&/r 
t2 

It  follows  that  k  =  — 2_- 

t  +  npd 

This  may  also  be  written 

2  n  dA  +  Bt2 


2nA  +  2Bt 

To  facilitate  the  solution  of  problems  the  following  curves, 
Fig.  224,  give  the  values  of  j  and  k  for  varying  values  of  p  and 

-  and  also  the  ratios  of  *£  for  various  values  of  k.    These  values 

/c 

have  been  calculated  with  n  —  15,  this  being  a  very  commonly 
assumed  value  for  rock  concrete.  For  other  values  of  n  similar 
curves  may  be  drawn  or  the  formulae  used. 


Single  Reinforcement 
MS=A  x  f8  x  j.d 


P  202 


FIG.  224. 


REINFORCED   CONCRETE  203 

To  findy  if  -  =  0.25;  from  0.25,  at  a  on  the  upper  scale,  drop 
d 

a  perpendicular  until  it  cuts  the  curve  corresponding  to  the  steel 
ratio  of,  say,  0.003  m  ^>  project  the  point  b  horizontally  until  it 
cuts  the  vertical  scale  in  c.  Here  the  value  of  j  is  given  as  0.914. 
To  find  k  continue  the  vertical  a-b  until  it  cuts  the  curve  of  k 
corresponding  to  a  steel  ratio  of  0.003  m  d.  This  point  d  pro- 
jected horizontally  cuts  the  vertical  scale  in  e  or  k  =  0.262. 

To  find  the  ratio  of  j  produce  the  value  of  k  horizontally  until 

Jc 

it  cuts  the  curve  of  j  in  /.    The  vertical  projection  of  /  cuts  the 

Jc 

scale  of  ratios  of  •'-j-  in  e  or  j  =  42 .3 . 

Jc  Jc 

An  example  will  further  illustrate  the  uses  of  the  formulae  and 
curves. 


[       •/»                T 

T~ 

A                      i 

t    f 

k 

t 

FIG.  225. 

Given  n  =  15,  fe  =  600,  /.  =  15,000,  and  the  following  section, 
Fig.  225.     Area  of  bars  =  4  X  0.766  =  3.06  sq.  ins. 

P  =  -7^~: —  =  0.0027. 
60  X  19 

Consulting  the  curves, 

j  =  0.918,     k  =  0.252,     •«  =  44.5. 

Jc 

The  resisting  moment  then  is 

Ma  =  A  XfaXj-d 
or 

Ma  =  3.06  X  15,000  X  0.918  X  19  =  801,000  in.  Ibs. 


204  GRAPHICS  AND   STRUCTURAL  DESIGN 

The  fiber  stress  in  the  concrete  is 

fa  I5,OOO  s 

7  =  44-5  =  -^7—      or    fe  =  330  Ibs.  per  sq.  in. 

Jc  Jc 

SLABS 

In  using  the  formulae  and  table  for  slabs  it  is  customary  to 
assume  a  width  of  slab  of  12  ins.;  thus  assuming  a  slab  whose 
span  is  6  ft.  and  whose  total  dead  and  live  load  is  200  Ibs.  per 
sq.  ft.,  taking  the  bending  moment  at 

Wl 
M= — ,  w  =  15,^.  =  600  Ibs.  per  sq.  in.,  p8— 12,000  Ibs.  per  sq.  in., 

the  thickness  of  the  slab  and  area  of  reinforcing  steel  can  be  found 
by  the  following  calculations: 

,,      Wl      (200  X  6)  X  6  X  12      Q,      .     „ 
M  =  —  =  —       — =  8640  in.  Ibs. 


10 

12,000 


=  I-33- 


15  X  600 

From  the  table  the  values  corresponding  to  z  =  1.33  are  k  = 
°-433>y  =  °-855>  k  -j  =  0.370  and  p  =  o.on;  substituting  these 
values  in  equation  (4)  gives 

,      J      2  M  c  J      2  X  8640 

d  =  V 


x  600x0.37      '        ' 

The  area  of  the  reinforcement  then  is  given  by  equation  (3) 
or  by  taking  b  -d  -p  =  12  X  2.55  X  o.on  =  0.337  scl-  m- 

If  f  in.  is  allowed  below  the  center  of  the  reinforcing  steel  the 
slab  thickness  will  be  2.  55  +0.75—3.  25  ins. 

BENDING  MOMENTS 

Owing  to  the  monolithic  construction  of  reinforced  concrete 
the  bending  moments  on  beams  are  frequently  estimated  as  lower 
than  those  occurring  on  similarly  loaded  but  merely  supported 
beams.  Thus  for  continuous  slabs  the  moments  are  commonly 

Wl 

taken  as  M  =  —  ,  at  the  middle  of  the  span. 
10 


REINFORCED   CONCRETE 


205 


Square  slabs  reinforced  in  both  directions  and  supported  on 
four  sides  have  the  moments  at  the  center  of  the  span  in  one 

Wl 

direction  taken  as  M  =  — .     Beams  and  girders  although  some- 

20 

Wl 

times  figured  as  fixed  or  continuous  beams  with  M  =  —  are  more 

10 

Wl 

commonly  taken  as  supported  beams  with  M  =  — .     In  the 

8 

above,  W  =  the  total  load  on  the  span  in  pounds  and  L  =  the 
span  in  inches. 

BEAMS  WITH  DOUBLE  REINFORCEMENT 

In  the  preceding  discussions  reinforcing  steel  has  been  assumed 
in  the  tension  side  of  the  beam.    The  case  will  now  be  considered 


- -—. -.^-^rzzzr:  EL"—  ~zi~  i^i.~ ~  ZZ^TT — = 


FIG.  226. 

where  the  reinforcement  is  in  both  flanges.  This  condition  occurs 
regularly  in  ordinary  construction  where  some  of  the  reinforcing 
bars  in  the  tension  flange  are  bent  up  and  carried  over  the  support 
in  the  upper  flange.  In  this  case  the  beam  resists  a  negative 
moment  over  the  supports  which  creates  tension  in  the  upper 
flange  and  compression  in  the  lower  flange.  In  the  case  of  T 
beams  the  lower  flange  width  being  b  the  width  of  the  stem  is 
much  less  than  the  width  B  of  the  upper  flange,  and  unless  this 
narrow  flange  is  reinforced  for  compression  the  fiber  stress  pc 
might  easily  be  excessive.  The  following  theory  applies  to 
beams  with  double  reinforcement. 

The  nomenclature  is  that  previously  given. 

Assuming  the  compression  in  the  concrete  as  varying  propor- 


206  GRAPHICS  AND   STRUCTURAL  DESIGN 

tionally  to  the  distance  from  the  neutral  axis  and  that  the  con- 
crete carries  no  tension,  we  have,  in  Fig.  226, 

nfc         kd  nfc         kd 

Also  Axf.  =  c.'Xc  =  A'f,'  +  $fe-kd.b, 

from  which  it  follows  that 

n(A'  +  A) 


and 


k  =      2»(#  +  0/>')+»*(/>  +  />')2  -  n  (p  +  /). 
To  find  the  lever  arm  of  the  couple  acting  in  the  beam 

A  Xf.  Xjd  =  ±fc.kd-b(d--\  +  A'f.'  (d  -  e), 

fc.kd.b(d--) 
-v  V         3/  ,  A'f.'(d-e). 

2Af,  A*f. 

Expressing  fcj  fsj  f,  and  d  in  terms  of  k,  <f>,  p  and  p'  gives 


J"  np  (i  -  *) 

The  relations  between  the  fiber  stresses  are 


,-.__.      f 

~AXjXd'  ~ 


and 


In  the  Engineering  News  of  August  22,  1912,  Mr.  Arthur  G. 
Hayden  gives  the  following  approximate  formulae  for  j  of  rein- 
forced-concrete  beams  with  double  reinforcements. 

j  =  0.93  +  2  P'  +  iooppf  —  4p  —  o.i0  —  2op'<f>. 

Mr.  Hayden  states  that  the  maximum  error  does  not  exceed 
if  per  cent  for  all  proportions  of  reinforcement  between  J  per 


REINFORCED   CONCRETE  207 

cent  and  2  per  cent,  and  that  for  ordinary  cases  the  error  is  in- 
appreciable. This  formula  assumes  that  n  =  15. 

The  following  example  will  illustrate  the  use  of  the  formulae. 

A  T  beam  in  which  d  =  21  ins.  and  b  =  12  ins.  is  reinforced 
over  the  supports  by  two  f  -in.  <£  bars  in  each  flange.  If  fa  = 
16,000  Ibs.  per  sq.  in.,  </>  =  ^  and  n  =  15,  what  moment  can 
the  beam  resist  at  the  supports  and  what  are  the  fiber  stresses 
PC  and  #/? 


2n(A  +  *A')d      \n(A+A'W     n(A+A') 

=   ~  f  ~ 


_ 

L         12         J 


12 


o.282fo.5o  -  2iL-J  -f  15  x  0.0048  (i.o  -  o.i)  (0.28  -  o.i) 

15  X  0.0048  (i  -  0.28) 
j  =  0.91     and    jd  =  21  X  0.91  =  19.11  ins. 

By  the  approximate  method  jd  =  18.97  ms- 

The  bending  moment,  allowing  /.  =  16,000  Ibs.,  is 

M  =  A  X/3  Xjd  =  1.2  X  16,000  X  19.11  =  366,912  in.  Ibs. 

f<  =  n(dM-kd)f'  =  IS  (21  '-5.85)    X  I 

,f     k  —  <f>  j.      0.28  —  o.io 
/,  =     __  ,  /.  =  -  -  -  —  X  16,000  =  4000  Ibs.  per  sq.  in. 

The  following  problems  are  intended  to  be  solved  by  means 
of  the  theory  just  given  and  it  is  recommended  that  they  be  first 


208 


GRAPHICS  AND  STRUCTURAL  DESIGN 


done  by  using  the  formulae  and  then  checked  by  the  assistance 
of  the  curves. 

Problem.  —  The  combined  live  and  dead  loads  upon  a  floor 
slab  are  140  Ibs.  per  sq.  ft. ;  assuming  fc  =  650  lbs.,/8  =  16,000  Ibs. 

per  sq.  in.,  -f  =  15,  and  that  M  =  — ,  determine  the  thickness 

JtLc  10 

of  the  slab  from  the  top  to  the  center  of  the  reinforcement  and 
the  spacing  of  f -in.  round  bars  if  the  span  of  the  slab  is  10  ft. 
8  ins. 

Problem.  —  Assume  a  continuous  girder  whose  span  is  25  ft. 
o  ins.,  carries  50,000  Ibs.,  d  =  23  ins.,  B  ~  100  ins.,  slab  thickness 

4^  ins.,  --r  =  15,  and  M  =  — '.    Find  the  fiber  stress  in  the  con- 
hc  10 

crete  and  steel  if  the  steel  reinforcement  consists  of  two  if -in. 
and  two  if-in.  round  bars. 

PARABOLIC  VARIATION  OF  STRESS  IN  CONCRETE 

In  the  preceding  discussion  of  reinforced-concrete  beams  it 
has  been  assumed  that  the  fiber  stress  in  the  concrete  above  the 


FIG.  227. 


FIG.  228. 


neutral  axis  varies  as  the  intercepts  in  the  triangle;  this  is  repre- 
sented by  Fig.  227. 

Some  designers  prefer  to  assume  that  this  stress  varies  as  the 
intercepts  in  the  parabola  as  shown  in  Fig.  228.  The  resultant 
flange  force  acts  at  a  distance  f  k  •  d  from  the  top  of  the  beam 
and  the  average  fiber  stress  is  f  fc.  This  treatment  is  commonly 
limited  to  rectangular  beams  as  it  becomes  too  complicated 
when  extended  to  T  beams.  The  following  formulae  are  derived 


REINFORCED  CONCRETE  209 

similarly  to  those  previously  determined  and  apply  to  rectan- 
gular beams  only.     The  symbols  are  those  previously  used. 


j      T  / 

and  *-'- 


M.  =  A  Xf.Xjd, 
Mc  =  lfcXbX  kj  X  d\ 
Where  the  ratio  of  reinforcement  is  known  the  formulae  are: 


k  =  V%p.n  +  (%p-n)2-lp-n    and    j  =  i  -  f  k, 


M. 

J»~      .     t       .' 


The  difference  between  these  formulae  and  those  previously 
given  will  be  evident  upon  working  several  of  the  problems  by 
both  methods.  Considering  the  character  of  the  materials  used 
in  reinforced  concrete  it  is  a  question  whether  methods  intro- 
ducing any  greater  refinements  than  those  used  in  the  derivation 
of  the  first  formulae  are  warranted. 

Most  elaborate  tests  of  reinforced-concrete  beams  have  been 
made  at  the  Experiment  Station  of  the  University  of  Illinois  under 
the  direction  of  Prof.  Arthur  N.  Talbot,  beginning  in  1904  and 
reported  in  the  Bulletins  of  that  station  since  then.  Professor 
Talbot  has  deduced  formulae  and  offers  suggestions  for  reinforced- 
concrete  design  as  the  results  of  the  tests ;  this  data  can  be  readily 
obtained  by  reference  to  the  above  bulletins. 

HORIZONTAL  SHEAR 

It  is  shown  in  the  section  on  problems  (see  problems  65  and 
68)  that  ordinarily  in  steel  beams  of  usual  sections  horizontal 
shear  is  of  minor  importance,  while  more  careful  consideration 
must  be  taken  of  it  in  timber  beams,  particularly  for  those  of 
short  span.  In  reinforced  concrete  it  is  of  still  greater  impor- 
tance and  generally  the  webs  must  be  carefully  reinforced  with 


2IO 


GRAPHICS  AND   STRUCTURAL  DESIGN 


steel  to  resist  this  shear.  Taking  section  2-2  in  Fig.  229  the 
horizontal  shearing  force  will  increase  from  zero  at  the  upper 
fibers  to  a  maximum  at  the  neutral  axis  i-i ;  this  force  F2  is  then 
transferred  to  the  horizontal  reinforcing  steel.  Assuming  that  the 
bending  at  3-3  exceeds  that  at  2-2  the  horizontal  shearing  force 


FIG.  229. 

F3  at  3-3  will  exceed  F2,  and  if  b  is  the  width  of  the  beam  at 

the  neutral  axis  the  unit  shearing  force  on  the  section  b  X  i  is 
p  p 

v  =  -r— 2 .      Equating  the  moments  V  X  i  =  (F9  -  F2)  j  •  d 


b  X  i 
or  F3  -  F 


and  v 


V 

b-j-d 
It  has  been  previously  shown  that  j  •  d  ~  f  d,  hence 

V 
0.85  b  -d 

It  should  be  noted  that  the  reasoning  is  identical  with  that 
used  in  determining  the  flange  riveting  in  plate  girders. 

The  design  is  sometimes  made  by  assuming  that  the  horizontal 
shear  is  carried  partly  by  the  concrete  and  any  deficiency  in  the 
concrete  is  supplied  by  the  addition  of  steel  stirrups.  These  are 
small  bars  or  strips  commonly  bent  into  U  or  V  shapes  and  placed 
vertically  in  the  web.  They  are  usually  secured  in  the  flange 
and  are  run  under  the  horizontal  reinforcing  bars  to  hold  them 
in  place. 

Let  vc  be  the  allowable  unit  working  shear  in  the  concrete;  then 
the  deficiency  per  square  inch  is  v  —  vc. 

If  R  =  the  value  in  pounds  of  the  stirrup  in  shear, 

S  =  stirrup  spacing  in  inches  at  the  section  where  the 
vertical  shear  is  V. 


REINFORCED   CONCRETE  211 

If  the  stirrup  carries  the  entire  shear,  then 

c      0.85  XdxR 
V 

If  the  stirrups  take  only  that  portion  of  the  shear  that  the  con- 
crete is  unable  to  carry,  then 

„  =        0.85  XdXR 
~  7-0.85  Xvc-b-d' 

It  is  noted  from  both  these  formulae  that  the  spacing  S  will  be 
closer  the  greater  the  value  of  V '.  It  therefore  follows,  as  in  the 
case  of  flange  riveting  of  plate  girders,  page  149,  that  the  stirrups 
will  be  spaced  closer  at  the  supports.  To  be  effective  the  stirrup 

spacing  should  not  exceed  — 

Graphical   Determination    of    Stirrup    Spacing.  —  Beams    or 
girders  of  a  constant  depth  d  carrying  a  uniform  load  may  be 


/;  i  I    ;         r        i    !  j  \ 

Cjn  o  p q  A  jD 

FlG.  231. 

considered  as  follows.  In  Fig.  230  lay  off  to  some  scale  the 
horizontal  line  CD  representing  the  span.  At  its  center  A ,  erect 
a  perpendicular  AB  representing  to  scale  the  flange  force  acting 
in  the  reinforcing  steel  or  in  the  concrete  flange  at  the  center  of 
the  span.  This  flange  force  will  equal  the  area  of  the  steel  multi- 
plied by  the  unit  fiber  stress  induced  in  it  by  the  loading.  This 
fiber  stress  can  be  found  by  the  formulae  previously  given  for 
beams. 

If  in  a  beam  A  =3.06  sq.  ins.,  and  fa  =  15,000  Ibs.  per  sq. 
in.,  the  flange  force  then  is  F  =  3.06  X  15,000  =  45,900  Ibs. 
AB  equals  45,900  Ibs.  and  the  change  of  flange  force  across 


212 


GRAPHICS  AND   STRUCTURAL  DESIGN 


the  girder  may  be  represented  by  the  parabola  CBD.  In-  the 
case  of  a  supported  span  the  flange  force  will  be  zero  at  the  sup- 
ports, while  in  a  restrained  beam  the  flange  force  will  go  through 
zero  at  some  points  E  and  F  shown  on  the  dotted  line  C1  —  D1, 
which  has  been  raised  from  the  position  C  —  D  by  the  restraint  on 
the  beam.  The  flange  force  at  the  supports  is  now  C  —  C1  instead 
of  zero  and  is  of  the  opposite  character  to  the  force  B-G  at  the 
center.  The  range  of  flange  force  is  still  A-B  being  CC1  +  BG. 
Now  if  AB  (Fig.  231)  is  divided  into  one-half  the  number  of 
parts  that  it  is  thought  advisable  to  use  stirrups  in  the  span,  AB 
is  here  divided  into  five  equal  parts,  and  lines  parallel  to  CD 
are  drawn  through  these  points,  i,  j,  k  and  /,  the  spaces  Cn,  no, 
op,  pq  and  qA  are  the  sections  in  each  of  which  a  stirrup  must 
be  placed.  The  value  of  each  stirrup  in  shear  must  equal  Ai. 


c\nop  q 


FIG.  232.  FIG.  233. 

If  it  is  desired  to  allow  the  concrete  to  carry  a  part  of  the  hori- 
zontal shear  it  may  be  allowed  for  by  laying  off  A  E  in  Fig.  232 
to  the  same  scale  as  AB  and  equal  to  the  product  of  the  allow- 
able unit  shear  on  the  concrete,  the  width  of  the  beam  (width  of 
stem  if  it  is  a  T  beam)  and  one-half  of  the  span,  the  first  in 
pounds  per  square  inch,  the  last  two  in  inches.  Draw  the  line 
CE  and  as  BE  represents  the  total  shear  that  is  to  be  cared  for 
by  the  stirrups  in  the  half  span,  divide  BE  into  the  number  of 
equal  parts  that  there  are  desired  stirrups  in  the  half  span.  The 
spaces  within  which  the  stirrups  must  be  placed  are  m,  «,  o,  p 
and  q,  being  located  by  the  intersections  of  the  parabola  with 
the  lines  through  i,j,  k  and  /  and  drawn  parallel  to  CE.  In  the 

best  practice  the  stirrups  are  spaced  a  distance  not  exceeding-- 

2 


REINFORCED   CONCRETE  213 

Tests  of  reinforced-concrete  beams  exhibit  failures  along  lines 
a-a  as  shown  in  Fig.  233.  This  illustrates  failure  by  diagonal 
tension.  If  we  consider  any  point  in  the  body  of  the  beam,  it 
will  be  acted  on  by  a  unit  horizontal  tension  or  compression,  a 
unit  horizontal  shear  and  a  unit  vertical  shear.  The  resulting 
maximum  tensile  or  compressive  stress  is 

A.  =  */,  +  ^i/,2  +  f2, 

where       ftm  =  unit  maximum  diagonal  tension, 
ft    =  unit  horizontal  tension, 
v     =  unit  shear,  vertical  or  horizontal. 

The  direction  of  the  maximum  tension  is  given  by 

2  V 

tang  26  =  — —  > 

h 

where  ft  =  o,  0  =  45  degrees.     It  therefore  follows  that  along 
the  neutral  axis  and  on  the  tension  side  of  this  axis  that  6  =  45 
degrees. 
.  Where  0  =  45  degrees,  ft  =  o  and  ftm  =  v. 

It  is  seen  then  that  the  maximum  diagonal  tension  acts  at  an 
angle  of  45  degrees  with  the  neutral  axis  and  that  the  maximum 
unit  diagonal  tension  equals  the  unit  shear  at  that  point. 

'--'-iTS' 

v  =  unit  vertical  shear. 

b  =  width  of  beam  if  rectangular.     Use  b\,  the  width  of  the 

stem  if  a  T  beam. 
V  =  total  vertical  shear  on  section. 

This  formula  applies  for  the  worst  condition,  namely,  that  no 
tension  is  carried  by  the  concrete.  When  it  is  desired  to  allow  for 
the  concrete  the  numerator  of  the  fraction  becomes  V  —  vc  •  b  •  d 
for  rectangular  beams  and  V  —  vc  •  b\  •  d  f or  T  beams.  Here 
vc  =  the  allowable  unit  shear  on  the  concrete. 

Let  T  =  the  total  allowable  tension  in  a  diagonal  bar,  in 
pounds.  If  the  bar  makes  an  angle  of  45  degrees  with  the  hori- 
zontal, the  spacing  s  along  the  horizontal  in  the  first  case  is  s  = 


214  GRAPHICS  AND   STRUCTURAL  DESIGN 

rr\  7 

*- 


while,  when  the  allowance  is  made  for  the  portion 


of  the  tension  carried  by  the  concrete, 

=  1.41  XT  Xd 
:  V-ve-b-d' 

The  factor  1.41  is  introduced  since  the  diagonal  tension  is 
measured  along  a  45  -degree  line,  while  the  spacing  is  laid  off  on 
a  horizontal  line. 

To  be  effective  for  web  reinforcement  the  spacing  of  diagonal 
bars  should  not  exceed  s  =  d. 

BOND  STRESS 

The  diagram,  Fig.  234,  shows  how  the  flange  force  varies  from 
the  supports  to  the  middle  of  the  span  for  a  uniformly  dis- 
tributed load  upon  a  beam  of  constant  depth.  Evidently  this 
change  of  force  can  only  occur  by  the  flange  transferring  force 

to  the  web,  as  was  done  in  the 
case  of  the  plate  girder  through 
the  flange  riveting.  AB  repre- 
sents the  maximum  flange  force. 
Passing  from  AB  to  EF  the  flange 
force  has  been  reduced  by  the 
amount  GF  which  has  been  trans- 
ferred  to  the  web.  In  the  case 


Q 


FlG>  234< 


of  reinforcing  steel  the  connection  between  the  steel  and  the 
concrete  must  be  strong  enough  to  carry  the  force  GF.  This 
may  be  accomplished  either  by  the  adhesion  between  the  con- 
crete and  smooth  or  plain  bars,  or  corrugated  or  twisted  bars 
may  be  used  to  increase  the  strength  of  the  bonding. 
Considering  plain  round  bars  and  letting 

u  =  unit  bonding  stress,  pounds  per  square  inch, 

d  =  diameter  of  the  bar,  inches 

/  =  length  of  bar,  inches, 


then, 


u  X  -  =  —  X//, 
24 


from  which- 
5 


2  u 


REINFORCED    CONCRETE  215 

This  means  that  to  develop  the  full  tensile  strength  of  a  round 

bar  its  total  length  must  be  -"-  times  the  diameter  of  the  bar. 

2  -  u 

This  relation  applies  also  to  square  bars.  Illustrating  this  with 
the  |-in.  square  bars  used  in  a  former  example,  and  allowing 
60  Ibs.  per  sq.  in.  for  the  working  bond  stress,  we  have 

/        /,          ;      0.875  X  16,000 

-  =  -^-or  /  =  -  -  =  116.5  ins- 

d      2  •  u  2  X  60 

This  means  that  under  the  given  conditions  to  develop  the  full 
strength  of  the  bar,  it  must  have  a  total  length  of  at  least  116.5 
ins.  and  must  be  placed  to  have  a  length  of  not  less  than  58^  ins. 
on  each  side  of  the  center  of  the  span. 

The  bond  stress  varies  along  the  bars;  in  fact,  the  increment 
of  change  is  that  of  the  flange  force,  and,  similarly  to  the  change 
in  horizontal  shear,  we  have 

V  X  I  =  (F3  -  Fjj.d,       :.      F,-F2  =  2— 

J  -d 


The  unit  bonding  stress  = 


F,-F2 


u  = 


surface  of  bars  for  unit  length 
V 


j  •  d  (2  circumference  of  bars) 


LENGTHS  OF  REINFORCING  RODS 

The  lengths  of  the  reinforcing  bars  besides  being  limited  by 
the  bond  stress,  as  just  explained,  will  also  vary  in  length  for 
the  same  reason  the  flange  plates  of  girders  do.  That  is,  as  the 
flange  force  decreases  from  the  center  towards  the  supports  the 
reinforcing  area  can  be  reduced.  In  Fig.  235,  CD  is  the  span 
and  AB  represents  the  total  reinforcing  area,  A,  at  the  center  of 
the  span.  If  4  bars  are  used,  the  line  AB  is  divided  into  4 
equal  parts,  each  division  representing  the  area  of  one  bar;  the 
lengths  of  the  several  rectangles,  determined  by  the  intersections 
of  the  horizontal  lines  with  the  parabola  CBD,  give  the  respec- 
tive lengths  of  the  bars.  In  practice  the  bars  instead  of  being 


2l6  GRAPHICS  AND   STRUCTURAL  DESIGN 

cut  off  are  frequently  bent  up,  run  to  the  top  and  then  to  the 
end  of  the  beam,  as  in  Fig.  236,  thus  reinforcing  the  upper  flange 
of  the  beam  over  the  support,  where,  owing  to  the  monolithic 
character  of  a  concrete  beam,  there  is  restraint  and  consequent 
tension.  The  bars  are  usually  bent  up  in  pairs.  The  area  run 
over  the  supports  generally  exceeds  25  per  cent  of  the  gross 
area  A\  the  remainder  of  the  bars  are  continued  through  the 


J  |p  -----  Span  ----- 

FIG.  235.  FIG.  236. 

lower  flange  and  over  the  supports.     If  it  is  preferred  to  calculate 
the  length  of  the  bars  the  following  formula  may  be  used, 


where       /  =  length  of  bar  in  lower  flange  between  bends,  inches. 

a  =  area  in  square  inches  of  bars,  including  all  shorter 

bars  and  the  bar  whose  length  is  desired.     In 

Fig.  235,  if  the  length  of  bar  No.  3  is  desired, 

a  =  area  of  bars  Nos.  i,  2  and  3. 

A  =  area  in  square  inches  of  the  total  reinforcing  at  the 

center  of  the  span. 
L  =  span  of  beam,  inches. 

DESIGN  OF  A  T  BEAM 

A  T  beam  spans  20  ft.  o  ins.  ;  the  beams  are  spaced  10  ft.  6  ins. 
and  carry  a  uniform  dead  load  of  30,000  Ibs.,  while  the  uniform 
live  load  is  75,000  Ibs. 

The  slab  thickness  is  7.5  ins.,  d  —  29  ins.,  and  the  width  of  the 
stem  61  is  20  ins.  Assume  the  width  of  slab  acting  as  flange  at 


REINFORCED   CONCRETE  217 

80  ins.     The  reinforcing  bars  are  four  if-in.  round  bars  and  five 
i^-in.  round  bars. 
The  total  area  of  the  steel  is 

Four  ig-in.  round  bars=  4  X  0.99  =    3.96  sq.  ins. 
Five  ij-in.  round  bars  =  5  X  1.23  =    6.15  sq.  ins. 

Total  10.  ii  sq.  ins. 

These  bars  to  be  spaced  not  less  than  i\  times  their  diameters. 
From  the  dimensions  given  we  have 


d      29 

IO.II 


=  ao°44- 

From  the  curves  we  find  that  k  =  0.305  and  that  j  =  0.90, 
also  y  =  34. 

Wl 

The  bending  moment  on  the  beam  being  assumed  as  M  —  —  » 

8 

,,         105,000  X  2O  X  12  -11 

M  =  —**-  -  =  3,  150,000  in.  Ibs. 

o 


K/T         A     f      -     i     f  >, 

M  =  A.f.  -j  •</;/.=  IOIio9QX29  =  ",ooo  Ibs.  per  sq.  m. 

Bent  Bars.  —  Assuming  that  four  i^-in.  round  bars  are  placed 
above  the  five  i  J-in.  round  bars  and  that  these  four  bars  are  bent 
alike,  the  length  of  the  four  bars  between  the  bends  will  be  . 


—  =  240  V  =  ISO  ms. 

A  r  io.ii 

Had  it  been  desired  to  bend  up  bars  i  and  2  before  bars  3  and  4 
were  bent  their  lengths  /  would  have  been 

r4/a  4/1.98 

l\  —  L  \  —  =  240  V  —       =  1 06  ins. 
v  A  v  io.li 

The  minimum  length  of  the  bars  to  develop  the  proper  bonding 
strength  is  given  by  the  formula 

/  _  g  X/a  _  1.125  X  12,000 


oo  ins. 
2  Xu  2  X75. 


2l8  GRAPHICS  AND  STRUCTURAL  DESIGN 

It  is  seen  that  the  106  ins.  required  for  flange  strength  provides 
ample  bonding  strength. 

WEB  REINFORCEMENT 

The  maximum  shear  at  the  center  of  the  span  will  occur  when 
the  beam  is  covered  with  the  live  load  from  its  center  to  one  of 
the  supports.  The  dead  load  being  30,000  Ibs.  the  live  load  is 
105,000  —  30,000  =  75,000  Ibs. 

The  load  on  one-half  of  the  span  then  is  37,500  Ibs.  The  live- 
load  reaction  (Fig.  237)  is 

=  9375  lbs. 


The  maximum  shear  diagram  is  then  given  by  Fig.  238. 

The  straight-line  approximation,  ab,  is  commonly  used. 

Figure  239  represents  a  reinforced-concrete  T  beam  from  the 
center  of  the  span  to  the  right  support. 

From  the  point  i,  the  intersection  of  the  center  line  of  the  span 
and  the  neutral  axis  of  the  beam  section,  draw  ij,  making  an 
angle  of  45  degrees  with  the  horizontal  axis  of  the  beam.  The 
unit  shear  at  any  point  below  the  axis  equals  the  unit  diagonal 
tension;  at  the  center  of  the  span  this  is 

.  ,         7  =  =  16.15  lbs.  per  sq.  in. 

biXd-     20X29 

A  line  ih  is  laid  off  from  i  at  right  angles  to  ij  and  to  a  length 
representing  16  lbs.  Through  the  point  k,  at  the  junction  of  the 
neutral  axis  and  the  line  of  reaction  at  the  right  support,  dr&wjk 
at  right  angles  to  ij.  The  unit  end  shear  is 

aife.J2iSS2.ita. 

bi  X  d      20X29 

This  value  is  laid  off  to  scale  in  jb.  hb  may  be  joined  by  a 
straight  line  or  the  true  curve  may  be  used  as  explained  in  Fig. 
238.  The  former  is  done  here,  it  being  as  accurate  as  the  prob- 
lem warrants.  If  it  is  assumed  that  the  concrete  can  carry 


REINFORCED   CONCRETE 


219 


30  Ibs.  per  sq.  in.  lay  off  ja  to  scale  representing  30  Ibs.  and  draw 
the  line  ag  parallel  to  ij.    The  bars  are  bent  after  their  lengths 

FIG.  237. 


|* 10  ft.O H  37500  #      Live  Load 

-1 ' L 

j«j 20ft-0" J 


ao'o- >|Rs. 

FIG.  238. 


in  the  lower  flange  exceed  the  distances  previously  calculated  as 
necessary,  one-half  these  lengths  being  used  as  the  lengths  are 


DIAGRAM  OF  MAXIMUM  SHEARS. 

FIG.  239. 


measured  from  the  center  line  c-c.    The  area  of  the  trapezoid 
abdc  multiplied  by  b\  gives  the  load  carried  by  the  rods  3  and  4. 


220  GRAPHICS  AND   STRUCTURAL  DESIGN 

The  mean  ordinate  of  the  trapezoid  is 

cd  -f  ab  _  42  -t-  60  _ 

—       —   -si- 

ac  =  19  ins.     The  area  abdc  1351X19  =  970  Ibs. 

The  total  load  on  the  bars  3  and  4  =  970  X  &i  =  970  X  20  = 
19,400  Ibs.  Hence  the  unit  tension  in  the  bars  is 

19400    =  9800  Ibs.  per  sq.  in. 
2  X  0.99 

The  tension  in  the  bars  i  and  2  will  be  considerably  less,  but 
this  is  made  necessary,  as  the  spacing  must  be  less  than  d,  or 
29  ins.,  and  the  length  of  these  bars  in  the  lower  flange  measured 
from  the  center  line  of  the  span  must  exceed  53  ins.  The  bent 
rods  are  ordinarily  assumed  as  acting  approximately  through  the 
center  of  gravity  of  the  trapezoid  representing  the  load  they 
carry;  thus  bars  3  and  4  should  pass  approximately  the  center  of 
gravity  of  the  trapezoid  cbda. 

Although  the  discussion  errs  on  the  side  of  safety  it  should  be 
remembered  that,  the  diagram  being  one  of  maximum  shears,  the 
shears  given  are  not  in  existence  at  the  same  time.  There  is  still 
a  diagonal  tension  efg  not  cared  for;  this  will  be  provided  for 
with  vertical  stirrups  and  the  whole  beam  will  be  strengthened 
by  their  insertion  across  the  span.  The  vertical  stirrups  are 

also    useful    in    holding    the    horizontal   bars  in 

place. 

Had  the  horizontal  shear  been  assumed  as  being 

resisted  by  the  vertical  stirrups  and  these  stirrups 

taken  as  ^-in.  round  bars  bent  as  shown  in  Fig. 

240,  making  four  sections  of  the  rod  that  would 
FIG.  240.  nave  to  be  sheared,  and  allowing  10,000  Ibs.  per 
sq.  in.  as  the  permissible  shear  fiber  stress  the  stirrup  value 
would  be  R  =  4  X  0.196  X  10,000  =  7840  Ibs.  The  vertical 
shears  are  given  by  the  diagram  of  maximum  shear  (Fig.  238), 
and  have  the  following  values  at  the  several  distances  from  the 
center  of  the  span. 


REINFORCED    CONCRETE  221 

Distance  from  center  Max.  shear, 

of  span,  feet.  pounds. 

8 43,875 

6 35,250 

4-.  •  • 27,ooo 

2 iS.OOO 

The  minimum  spacing  will  be  at  the  supports  and  will  be 
=  °-85 dxR  =  0.85X29X7840 .ns 

~  7-0.85 xvcxbiXd  ~~  52, 500 -(0.85x50x20x29)     711 

The  spacing  8  ft.  from  the  center  of  the  span  is 

0.85  X  29  X  7840 

s  =  -  -  ~  10  ins. 

43,875  -  (°-85  X  50  X  20  X  29) 

The  spacing  6  ft.  from  the  center  of  the  span  is 

s  = 0.85  X  29  X  7840 _  ^ 

35,250  -  (0.85  X  50  X  20  X  29) 

The  spacing  will  be  made,  starting  at  the  piers  and  running 
towards  the  center,  2  spaces  at  8  ins.,  2  spaces  at  10  ins.,  and 
from  there  to  the  center  the  spacing  will  be  1 2  ins.  This  makes 

all  the  spacing  less  than  -,  or—  =  14.5  ins.,  which  would  be  the 

2  2 

maximum  for  the  best  practice.  The  vertical  stirrups  might 
have  been  made  much  lighter  considering  what  the  bent  rods  do 
to  strengthen  the  web,  but  in  this  case  the  vertical  stirrups  have 
been  figured  as  resisting  the  entire  horizontal  shear,  the  bent  rods 
will  strengthen  the  web  and  will  strengthen  the  beam  to  resist 
bending  by  resisting  a  negative  moment  over  the  supports. 

COLUMNS 

Concrete  columns  are  reinforced  in  one  or  both  of  two  ways: 
(i)  by  steel  rods  paralleling  the  axis  of  the  column  and  (2)  by 
spirally  wound  metal  bands.  In  the  first  case  the  metal  shares 
the  load  with  the  concrete,  while  in  the  second  case  the  bands 
strengthen  the  concrete  by  preventing  lateral  expansion  which 
is  assumed  as  occurring  when  the  column  is  shortened  along  its 
axis  by  the  load.  As  it  is  customary  to  tie  the  longitudinal 


222  GRAPHICS  AND  STRUCTURAL  DESIGN 

reinforcements  in  place  with  steel  wires  or  bands  it  virtually 
amounts  to  combining  both  forms  of  reinforcements. 

The  following  is  the  usual  discussion  of  the  strength  of  columns 
having  longitudinal  reinforcements. 

The  nomenclature  is 

P  =  total  load  on  column,  pounds. 

Ac  —  area  of  concrete   inside  reinforcements,   not  including 

steel,  square  inches. 
At  =  area  of  steel,  square  inches. 

A  =  Aa  +  Ac. 
A 
p  =  ratio,  -^- 

If  the  column  with  longitudinal  reinforcement  is  loaded,  its 
length  will  be  altered,  the  steel  and  concrete  being  shortened  the 
same  amount.  The  unit  reduction  is 

A=A  =  £. 
E.     E.' 

TJ* 

from  which  fa  =  fcX-=r  =  n  X  fc. 

&c 

The  total  load  carried  by  the  column  then  is 

P  =  Aa ./.  +  Ac  .fe  =  Aa .  nfe  +  Ac  *fc, 
and   P  =  (A.  - n  +  Ae)fe-,  also  P  =  A  -fe[i  +  p  (n  -  i)]. 

The  working  dimensions  of  the  columns  are  generally  assumed 
as  those  inclosed  by  the  longitudinal  or  band  reinforcements. 

The  proportions  of  concrete  columns  ordinarily  work  out  so 
large  that  their  ratio  of  length  to  smaller  dimensions  will  usually 
be  less  than  15,  thus  making  them  short  columns.  The  design 
may  be  safely  made  upon  the  basis  just  given,  even  where  length 
divided  by  the  smaller  side  reaches  25. 

The  usual  coating  of  concrete  is  commonly  put  on  the  column 
outside  the  reinforcing  steel  to  protect  it  and  act  as  fireproofing. 

A  few  of  the  numerous  forms  of  metal  reinforcements  for  con- 
crete are  shown  in  Figs.  241  to  247  inclusive.  Fig.  241  shows  a 


REINFORCED   CONCRETE 


223 


FIG.  241. 


FIG.  242. 


FIG.  243- 


FIG.  244. 


FIG.  245. 


FIG.  246. 


CROSS  SECTION  OF   BAR 


FIG.  247. 


224  GRAPHICS  AND  STRUCTURAL  DESIGN 

small  section  of  expanded  metal;  this  is  made  in  varying  sizes; 
the  smaller,  called  metal  lath,  is  intended  to  replace  wooden 
laths.  For  concrete  reinforcement  the  3 -in.  mesh  expanded  metal 
is  better.  It  is  made  in  weights  having  a  cross-sectional  area  of 
from  0.06  to  0.60  sq.  in.  per  ft.  of  width.  The  triangular-mesh 
steel-wire  reinforcement  illustrated  in  Fig.  242  serves  about  the 
same  purposes  as  the  expanded  metal.  These  reinforcements 
have  the  advantage  of  reinforcing  in  both  directions,  that  is, 
across  the  span  and  at  right  angles  to  it. 

Figure  243  is  the  Ransome  twisted  bar,  Fig.  244  is  the  Thatcher 
bar  and  Figs.  245  and  246  are  corrugated  bars.  These  bars  are 
designed  to  insure  proper  bonding  between  the  steel  and  the 
concrete. 

Figure  247  is  the  Kahn  trussed  bar;  flanges  along  the  side" of 
the  bar  are  bent  up  to  resist  the  horizontal  shear  in  the  web  of 
the  beam. 

The  Pittsburgh  Steel  Products  Company  make  a  reinforcing 
frame  consisting  of  steel  bars  through  the  lower  part  of  the  beam. 
To  these  bars  on  each  side  of  the  middle  of  the  span  are  welded 
shear  bars  bent  up  and  toward  the  adjacent  support  at  an  angle 
of  45  degrees. 

To  the  upper  end  of  these  shear  bars  are  welded  horizontal 
bars  which  run  back  over  the  support.  These  frames  are  made 
in  standard  sizes  and  the  reinforcement  in  the  upper  flange  over 
the  supports  is  said  in  all  cases  to  be  at  least  25  per  cent  of  the 
metal  in  the  lower  flange.  These  frames  are  built  up  for  certain 
spans  and  are  placed  completed  in  the  forms  about  which  the 
concrete  is  then  poured. 


CHAPTER  XV 
FOUNDATIONS 

THE  designer  recognizes  two  kinds  of  foundations,  those  for 
structures  and  others  for  machines.  The  function  of  the  former 
is  to  distribute  the  pressure  properly  upon  the  soil  and  where 
settlement  is  inevitable  to  design  the  foundations  that,  as  far 
as  possible,  such  settlement  shall  be  uniform  throughout  the 
structure.  It  may  also  maintain  the  structure  upright  against 
wind  or  other  forces  tending  to  upset  it,  as  in  the  case  of  a  chim- 
ney. In  machine  foundations  besides  these  functions  the  mass 
of  the  foundation  may  play  an  important  part  in  absorbing 
shock  or  may  even  prevent  the  motion  of  a  machine  when  acting 
on  other  machines  upon  foundations  external  to  its  foundation. 
Properly  balanced  rotary  converters  and  turbo-generators  re- 
quire only  rigid  support,  while  belt-driven  machines,  engines 
driving  rolling  mills,  mill  housings,  etc.,  must  be  anchored. 

With  modern  concrete  floors,  foundations  for  most  machine 
tools  may  be  dispensed  with,  the  machines  being  placed  where 
desired  and  rag  bolts  used  to  hold  the  machine  to  the 
floor. 

The  mass  of  a  foundation  to  limit  the  vibration  of  an  engine  or 
other  rapidly  moving  and  imperfectly  balanced  machine  within 
definite  limits  can  only  be  determined  when  complete  informa- 
tion of  the  design  of  the  engine  or  machine  is  available.  The 
determination  of  this  mass  therefore  is  properly  the  work  of  the 
engine  or  machine  designer  and  not  within  the  scope  of  this  book. 
Numerous  rules  of  thumb  have  been  used  in  determining  the 
minimum  weights  of  foundations.  One  of  these  is  to  make  the 
foundation  weigh  1.5  times  as  much  as  the  weight  of  the  engine 

225 


226  GRAPHICS  AND  STRUCTURAL  DESIGN 

or  machine.    Another  rule  credited  to  E.  W.  Roberts  is  that 
F  =  0.21  EVN,  where 

F  =  weight  of  the  foundation,  in  pounds. 
E  =  weight  of  the  engine,  in  pounds. 
N  =  R.P.M. 

The  wide  variations  in  these  two  rules  show  how  unsatisfactory 
such  formulae  are.  Vertical  engines  require  heavier  foundations 
than  horizontal  engines  of  the  same  capacity  and  speed. 

The  foundations  of  machines  subjected  to  much  shock  should 
be  kept  free  from  other  foundations  and  the  building. 

Although  considering  only  pressure  and  settlement  the  best 
foundation  bed  is  rock,  yet  vibrating  machines  placed  directly 
upon  the  rock  may  result  in  the  transfer  of  the  vibrations  through 
the  rock  for  considerable  distance.  The  usual  precaution  under 
such  circumstances  is  to  place  2  or  3  feet  of  sand  held  in  a  pocket 
in  the  rock;  this  may  be  either  a  natural  pocket  or  it  may  be 
made  of  concrete.  The  foundation  is  then  built  upon  this  cushion 
of  sand. 

In  setting  an  engine  or  important  machine  the  foundation  is 
brought  to  within  f  in.  to  ij  ins.  of  its  final  top.  The  machine 
is  then  lined  up  by  driving  metal  wedges  between  the  foundation 
and  machine  bed.  When  the  machine  is  properly  in  line  a  dam 
is  built  around  the  foundation  top  and  a  thin  grout  poured  in, 
filling  up  the  space  between  the  bed  plate  and  the  foundation. 
After  the  grout  hardens  it  firmly  secures  the  bed  and  the  founda- 
tion regardless  of  the  finish  of  the  engine  bed  or  foundation  top. 
When  the  grout  is  properly  set  the  wedges  may  be  removed 
and  the  spaces  left  by  them  filled  up  with  cement  mortar.  The 
foundation  bolts  are  preferably  located  and  held  in  place  during 
the  construction  of  the  foundation  by  templates  of  the  bed  plate. 
The  bolts  may  either  be  set  right  in  the  concrete  or  may  be 
placed  in  pipes  whose  inside  diameters  somewhat  exceed  the  bolt 
diameters  and  the  pipes  are  then  set  in  the  concrete.  The  bolts 
are  carried  by  the  templates;  the  space  between  the  top  of  the 


FOUNDATIONS 


227 


pipe  and  the  bolt  is  filled  with  burlap,  waste  or  paper,  to  prevent 
the  concrete  filling  the  clearance  space  between  the  pipe  and  the 
bolt. 

This  clearance  permits  some  adjustment  of  the  bolts  should 
the  bed  plate  disagree  with  the  template.  The  foundation  bolts 
should  be  run  well  down  into  the  foundation.  Frequently  the 
nuts  or  keys  are  placed  in  pockets  made  accessible  from  the  out- 
side of  the  foundation.  This  permits  of  the  ready  replacement 
of  a  bolt  should  it  be  found  to  be  necessary. 

In  some  cases  what  is  known  as  a  rust  joint  is  made  between 
the  foundation  and  the  bed  plate.  This  is  made  by  rusting  cast- 
iron  chips  together  with  sal-ammoniac.  Melted  sulphur  was 


D 


BOLT 

FIG.  248. 


KEY 

FIG.  249. 


WASHER 

FIG.  250. 


also  much  used  before  cement  grout  became  so  common.  The 
foundation  bolts  generally  have  the  end  above  the  foundation 
threaded  for  a  nut;  the  other  end  may  either  have  a  thread  and 
nut  or  be  slotted  for  a  key.  The  washers  in  the  foundation  are 
of  a  great  variety  including  scrap  channels,  angles  or  rails. 

Cast-iron  washers  may  also  be  used  but  of  whatever  kind  they 
should  be  large  to  distribute  the  pressure  well  over  the  concrete. 
Fig.  248  illustrates  a  slotted  bolt,  while  Fig.  249  shows  the  usual 
key  and  Fig.  250  the  cast-iron  washer.  In  other  cases  the  ad- 
justment of  the  bolt  is  provided  for  only  at  the  top  of  the  bolt 
for  a  distance  of  a  couple  of  feet.  This  is  done  by  using  a  short 
piece  of  pipe  similar  to  the  way  mentioned  before  or  by  setting 
a  wooden  casing  around  the  bolt  which  leaves  a  pocket  in  the 
finished  foundation  around  the  top  of  the  bolt.  The  foundation 


228 


GRAPHICS  AND   STRUCTURAL  DESIGN 


should  be  carried  to  proper  soil  and  always  below  the  frost  line. 
The  load  upon  the  soil  should  be  calculated  to  see  that  it  does 
not  exceed  that  permissible.  The  allowable  pressure  upon  the 
soil  is  commonly  given  by  the  building  codes  where  the  founda- 
tion is  within  a  city's  limits. 

It  may  frequently  be  had  by  obtaining  data  about  neighboring 
foundations.  The  following  table  from  "  Baker's  Masonry  "  is 
also  available  as  a  general  guide. 


Soil. 

Bearing  power, 
tons  per  sq.  ft. 

.  Soil. 

Bearing 
power,  tons 
per  sq.  ft. 

Rock,  thick  layers  

200  or  over 

Gravel  and  coarse  sand  .  . 

8-10 

Rock,  equal  to  best  brick. 
Clay   thick  beds,  dry 

15-20 
4-6 

Sand,  compact  
Sand,  clean   

4-6 
2—4 

Clay  thick  beds,  moder- 

Alluvial soils  

P5 

ately  dry  

2-4 

Clay,  soft 

1—2 

Where  the  engine  or  machine  is  an  important  one  and  the  soil 
of  doubtful  value,  piles  are  frequently  resorted  to.  The  subject 
of  piles  is  considered  under  building  foundations. 


H6le 


FIG.  251. 

Figure  251  illustrates  a  foundation  for  a  machine.  The  bolts 
have  been  set  in  pipes  and  run  through  to  pockets  accessible 
from  the  outside  so  that  the  bolts  could  be  replaced  if  necessary. 
This  foundation  shows  the  base  extended  to  reduce  the  pressure 
upon  the  soil  and  a  semicircular  hole  has  been  cut  through  the 
foundation  to  reduce  its  weight. 


FOUNDATIONS 


229 


In  another  class  of  foundations  for  machines  the  foundation 
requires  weight  to  prevent  the  overturning  of  the  machine. 
Fig.  252  illustrates  such  a  foundation  for  a  crane.  The  load  of 
33,000  Ibs.  is  to  be  carried  at  a  radius  of  33  ft.  The  frame  and 


425000* 


Force  Polygon 


Distribution  of 
Pressure  on  Soil 


FlG.  252. 


machinery  is  assumed  as  weighing  11,000  Ibs.  and  as  acting  at 
a  distance  of  6|  ft.  from  the  post  center.  The  foundation  is 
circular  and  has  been  assumed  as  weighing  425,000  Ibs.  The 
total  weight  on  the  foundation  then  will  be  the  sum  of  these  three 
amounts,  33,000  +  n,ooo  +  425,000  =  469,000  Ibs. 


230  GRAPHICS  AND  STRUCTURAL  DESIGN 

The  distance  of  the  line  of  action  of  the  resultant  of  these  forces 
can  be  found  by  taking  moments  about  the  vertical  axis  of  the 
post;  hence 

_  (33  X  33,000)  +  (6.5  X  1  1,  OOP)  . 

x  —  —  —  2.40  it. 

469,000 

The  maximum  pressure  should  now  be  determined  as  is  done 
for  a  chimney  (see  page  245).  The  foundation  has  been  assumed 

as  circular.     The  kern  radius  of  a  circle  is  —  •    Here  then  r  =  - 

8  8 

=  2.0  ft.     The  resultant,  therefore,  falls  outside  the  kern.     Re- 
ferring to  Fig.  270,  we  first  find 


then  a  =  0.355     and     W  =  <*peD2, 

W          460.000 

or  pe  =  —  —  =  —  —  =  5160  Ibs. 

aD2      0.355  X  i62 

Whether  or  not  this  maximum  pressure  is  permissible  will  de- 
pend upon  the  character  of  the  soil  or  nature  of  the  foundation 
under  the  section  considered.  The  curves  give  <£  as  41  per  cent 
and  the  neutral  axis  therefore  falls  16  X  0.41  =  6.55  ft.  to  the 
left  of  the  vertical  axis,  or  1.45  ft.  to  the  right  of  the  left  edge 
of  the  foundation  when  the  full  load  is  in  the  position  shown  in 
the  figure.  The  distribution  of  this  pressure  is  indicated  below 
the  assumed  foundation  bottom. 

BUILDING  FOUNDATIONS 

Since  settlement  of  foundations  is  bound  to  occur,  unless 
foundations  rest  on  bed  rock,  the  foundations  of  a  structure 
should  be  designed  as  far  as  possible  to  lead  to  uniform  settle- 
ment. The  effect  of  unequal  settlement  is  frequently  evident 
where  chimney  foundations  have  been  carried  into  wall  footings. 


FOUNDATIONS  231 

Here  the  settlement  of  the  chimney  may  cause  considerable 
cracking  of  the  wall. 

In  designing  foundations  it  is  considered  better  to  design  the 
areas  proportionally  to  the  dead  loads,  which  act  continuously, 
and  then  see  that  they  are  sufficiently  large  to  carry  the  combined 
dead  and  live  loads  without  bringing  excessive  pressure  upon  the 
soil.  The  dead  loads  acting  continuously  will  exert  a  greater 
effect  upon  the  settlement  of  the  foundation  than  live  loads 
which  may  act  but  infrequently  and  then  possibly  be  only  a 
small  part  of  the  assumed  live  loads. 

When  the  footings  are  proportioned  for  both  dead  and  live 
loads  only  such  portion  of  the  live  loads  should  be  considered 
as  may  be  assumed  as  acting  continuously.  It  is  also  of  prime 
importance  that  as  far  as  possible  the  center  of  pressure  on  the 
soil  shall  coincide  with  the  center  of  gravity  of  the  footing.  The 
building  codes  of  the  various  cities  give  the  requirements  of  foot- 
ings and  these  are  commonly  required  to  be  a  certain  enlarge- 
ment of  the  wall  upon  them.  Where  these  codes  are  not  available 
the  footings  can  be  designed  using  the  tables  of  allowable  pres- 
sures previously  given.  The  areas  should  be  enlarged  under 
pilasters  carrying  heavy  concentrations. 

Foundations  for  building  columns,  in  addition  to  resisting 
vertical  loads,  may  carry  the  horizontal  reactions  of  wind  loads, 
as  in  the  case  of  the  bent  for  the  steel-mill  building  designed  on 
page  118.  Here  there  will  be  two  cases,  one  (Fig.  253)  where 
the  column  is  assumed  as  hinged  and  then  the  horizontal  reac- 
tion acts  at  the  column  base;  the  other  (Fig.  254)  where  the 
column  is  considered  as  fixed  and  the  point  of  application  of  the 
horizontal  wind  reaction  is  taken  midway  between  the  foot  of 
the  knee  brace  and  the  column  base. 

The  base  is  commonly  a  rectangle.  Were  there  no  horizontal 
force  the  pressure  upon  the  foundation  would  be  uniformly  dis- 
tributed. The  wind  pressure  tends  to  increase  the  pressure  upon 
the  leeward  side  of  the  foundation  and  decrease  it  upon  the 
windward  side. 


232 


GRAPHICS  AND  STRUCTURAL  DESIGN 


As  long  as  the  resultant  pressure  R  cuts  the  bottom  of  the 

n 

foundation  at  a  distance  not  exceeding  —  from  the  center  of  the 

base  there  will  be  compression  over  the  entire  base.  R  is  the 
resultant  of  W  and  H,  where  W  =  load  carried  by  the  column 
or  wall  plus  the  weight  of  the  foundation  and  H  =  horizontal 
wind  reaction  on  the  column.  Since  neither  masonry  nor  the 
surfaces  of  the  foundation  and  soil  in  contact  can  be  in  tension, 


FIG.  254. 

having  R  pass  beyond  the  middle  third  will  reduce  the  area  of 
the  foundation  under  pressure  and  will  consequently  increase 
the  pressure  on  the  leeward  side,  see  Fig.  254. 

When  R  falls  in  the  middle  third  the  extreme  pressures  on  the 
edges  of  the  foundation  are  given  by 


here  b  =  width  of  the  base  at  right  angles  to  B. 


FOUNDATIONS 


233 


When  R  falls  beyond  the  middle  third  pe 


The  distance  x  may  be  found  graphically,  as  shown,  or  by 
moments,  since 

W        -H    h  =  H'h 

W 

In  the  second  case  (Fig.  254)  the  attachment  of  the  columns 
to  the  bases  must  be  sufficiently  strong  to  develop  the  bending 

moment  H  X  -  at  this  point. 

2 

The  more  general  discussion  of  the  kern  of  a  section  and  the 
distribution  of  pressure  upon  the  soil  is  given  under  chimney 
foundations,  page  253.  Where 
the  base  of  a  foundation  ex- 
tends beyond  the  main  shaft, 
Fig.  255,  the  portion  extended 
should  be  calculated  as  a  can- 
tilevered  beam  subjected  to  a 
uniform  load  pe.  This  is  com- 
monly stated  as  tons  or  pounds 
per  square  foot.  This  exten- 


• 


FIG.  255. 


i 


sion  e  is  given  by  e  =  4  d 


All  dimensions  in  inches. 


p    =  30  to  50  Ibs.  per  sq.  in.  in  concrete. 

pe  =  maximum  load  on  soil  in  pounds  per  square  foot. 

Figure  256  gives  a  foundation  standard  adopted  by  one 
company.  Capstone  —  i  Portland  cement,  2  sand,  4  blast  fur- 
nace slag.  Foundation  —  i  Portland  cement,  2  sand,  6  slag, 
minimum  bolt  diameter  —  i J  in.  —  pressure  on  metal  at  base 
of  column  not  over  10  tons  per  sq.  ft.  Pressure  on  capstone 
not  over  5  tons  per  sq.  ft.  Bearing  pressure  per  square  foot  on 
gravel  2.5  tons.  Bearing  pressure  per  square  foot  on  other  soil 
2  tons. 


234 


GRAPHICS  AND   STRUCTURAL  DESIGN 


Minimum  12 
to  suit  excava 


"A" 
so      : 

-t 


B"  determined  "by  30° 

FIG.  256. 


FIG.  257. 


Where  a  shallow  and  light  foundation  is  sufficient  a  grillage 
or  reinforced-concrete  footing  can  be  used.     The  beams  or  rails 

used  in  Fig.  257  may  be  cal- 

~*~a^:~  ^"~8     culated  for  the  load  given  by 

a  diagram  similar  to  that  of 
Fig.  255. 

The  extended  part  is  as- 
sumed as  carrying  a  uniform 
load  equal  to  the  length  e,  in 
feet,  multiplied  by  the  maxi- 
mum load  pe,  in  pounds  per 
square  foot. 

In  the  reinforced-concrete 
footing  (Fig.  258)  the  rein- 
forcement can  be  only  roughly 
estimated.  One  method  is  to 
consider  the  load  upon  the 
trapezoid  abed  and  assume  it 
as  carried  by  the  width  be. 
The  load  W  on  this  trapezoid 


FlG-  258- 


is 


;  the  distance  x  to  the  center  of  gravity  of  the 


trapezoid  is  x  = 


2^1  x  A 

-  BI  3 


FOUNDATIONS  235 

The  bending  moment  is  M  =  W  (A  —  x).  In  the  formula 
for  W,  all  dimensions  are  in  feet,  pe  =  Ibs.  per  sq.  ft.  and  W  is 
in  pounds.  In  the  formula  for  x  all  dimensions  are  in  inches. 
In  the  equation  for  M  all  the  terms  are  in  pounds  and  inches. 
Having  found  the  moment  M  the  design  is  readily  completed  as 
explained  under  reinforced  concrete  and  as  illustrated  by  the 
design  under  a  reinforced-concrete  chimney. 

PILES 

Where  the  sustaining  power  of  the  soil  is  inadequate  any  of 
the  preceding  foundations  may  be  placed  upon  piles.  Piles  were 
until  a  few  years  ago  entirely  of  timber,  preferably  white  oak, 
with  a  point  diameter  of  at  least  6  ins.  but  frequently  specified 
as  8  to  10  ins.  Their  lengths  will  vary  to  suit  the  particular 
location.  During  the  last  few  years  extensive  use  has  been  made 
of  concrete  piles.  Timber  piles  should  be  sharpened  at  the 
point,  and  are  frequently  shod  with  an  iron  band  or  point  to 
protect  them.  The  larger  end  is  cut  off  square  and  is  also  some- 
times protected  with  iron  caps,  hoops  or  bands.  The  brooming 
of  the  butt  of  a  pile  interferes  materially  with  driving  it,  so  that 
such  piles  should  be  trimmed  during  the  driving  to  facilitate  this 
operation.  The  final  or  test  blows  should  not  be  made  on  a 
broomed  pile. 

Piles  may  carry  their  load  through  friction  between  their  sides 
and  the  surrounding  soil,  or  they  may  be  driven  to  rock  when 
the  load  may  be  carried  largely  by  the  rock.  A  pile  may  there- 
fore fail  through  crushing  of  the  timber,  or,  where  a  pile  is  driven 
to  rock  or  its  equivalent  through  loose  soil,  it  may  fail  as  a 
column.  Specifications  limit  the  load  on  a  pile  to  from  40,000 
to  50,000  Ibs.,  or  600  Ibs.  per  sq.  in.  of  its  mean  section.  When 
a  possible  failure  as  a  column  is  considered  the  load  should  be 
calculated  for  a  column  whose  maximum  stress  does  not  exceed 
600  Ibs.  per  sq.  in. 

Timber  piles  when  completely  submerged  in  water  and  not 
exposed  to  the  toredo  (shipworm)  are  apparently  preserved 


236 


GRAPHICS  AND  STRUCTURAL  DESIGN 


indefinitely.  It  is  therefore  necessary  to  cut  timber  piles  below 
extreme  low  water.  A  bed  is  then  built  upon  the  tops  of  the 
piles  to  carry  the  upper  part  of  the  foundation  (see  Fig.  259). 
The  usual  spacing  of  the  piles  will  vary  from  two  to  four  feet  and 
upon  the  tops  of  the  piles  which  have  been  sawed  off  to  a  common 
level  is  placed  a  grillage  of  timbers  secured  to  the  piles  with 
metal  pins.  The  masonry  is  then  placed  upon  this  timber.  All 
timber  must  be  below  water.  Another  and  later  practice,  shown 
in  Fig.  260,  is  to  incase  the  tops  of  the  piles  in  a  bed  of  concrete. 
As  in  the  first  case  the  piles  are  cut  off  below  the  water  level. 


fe 


The  soil  around  the  piles  is  excavated  for  a  couple  of  feet  and  a 
bed  of  sand,  say,  12  ins.  thick,  is  put  in,  and  upon  this  is  laid  the 
concrete  whose  thickness  depends  upon  the  load  to  be  carried. 
This  last  method  firmly  incases  the  pile  tops  and  distributes  a 
part  of  the  load  upon  the  soil. 

Concrete  piles  have  an  advantage  over  timber  piles  in  not 
being  attacked  by  the  toredo  and  in  not  having  to  be  continu- 
ously submerged. 

There  are  several  types  of  patented  concrete  piles.  The 
Simplex  is  made  in  place  by  driving  a  hollow  tube  with  a  pointed 
end  the  entire  depth  required  and  then  filling  the  hole  with  con- 
crete as  the  tube  is  withdrawn.  In  the  Raymond  pile  a  thin 
tube  is  driven  from  which  a  collapsible  case  is  then  removed  and 
replaced  with  concrete.  Other  forms  of  reinforced  piles  are 


FOUNDATIONS  237 

made  on  the  ground  and  when  properly  aged  driven  with  a  pile 
driver. 

There  is  no  very  satisfactory  method  of  determining  the  bear- 
ing power  of  a  pile.  The  best  known,  where  the  pile  is  not  driven 
to  refusal,  is  the  Engineering  News  Pile  Formula.  This  formula 

2  -W  -h  ,  2  -W  -h  f 

is  P  =  -          -  for  drop-weight  hammers  and  P  =  -     — r—  for 

S+l  S  -f  TO 

steam  hammers. 

P  =  safe  load  in  tons,  2000  Ibs. 
W  =  weight  of  hammer  in  tons,  2000  Ibs. 
h  =  drop  of  hammer  in  feet. 
s  =  penetration  in  inches  due  to  the  last  blow. 


CHAPTER   XVI 


CHIMNEYS 

IN  the  design  of  masonry  chimneys  it  is  usual  to  assume  that 
the  masonry  cannot  resist  tensile  stresses.  If  the  weight  of  the 
column  above  the  section  i-i,  Fig.  261  (a), 
is  W  pounds  this  load  will  be  uniformly 
distributed  on  the  section  when  no  other 
forces  act  on  the  column  above  this  section. 
If  then  the  area  is  A  sq.  ins.  the  uniform 


-tr-* 


(«>      I.  ,    .  w 

fiber  stress  is  —  Ibs.  per  sq.  in.     Assuming 
A. 

a  force  F  acting  on  the  right  of  the  column, 
it  will  tend  to  reduce  the  fiber  stress  on  the 
right  and  increase  that  on  the  left  of  the 
section.  The  original  fiber  stress  /  is  com- 
pression. If  the  force  F  is  increased  suffi- 
ciently the  fiber  stress  on  the  right  would 
become  zero  while  that  on  the  left  would 
be  doubled,  as  is  indicated  in  Fig.  261  (d). 
The  unit  stress  would  vary  across  the  sec- 
tion as  shown  in  Fig.  261  (d),  being  as- 
sumed as  varying  uniformly.  The  stresses 
would  still  all  be  compression.  Under  these 
conditions  the  resultant  R  of  W  and  F  must 
act  through  a  point  a  distance  q  from  the 
center  of  gravity  of  the  section  i-i.  Should 
the  resultant  R  act  at  a  greater  distance 
from  the  center  of  gravity  of  the  section 
than  q,  which  would  be  due  to  a  still  further  increase  in  F,  then 
the  fiber  stress  on  the  right  would  pass  through  zero  and  be- 

238 


(fl 


FIG.  261. 


CHIMNEYS  239 

come  tension,  should  the  material  resist  tension,  and  the  fiber 
stress  on  the  left  would  be  increased  but  would  continue  to  be 
compression.  Should  the  material  not  be  able  to  resist  tension 
the  pressure  will  adjust  itself  to  the  left  until  the  resultant 
stress  R  coincides  with  the  point  of  intersection  of  the  resultant 
of  W  and  F  with  the  section  i-i.  When  the  extreme  fiber 
stress  on  the  left  exceeds  the  strength  of  the  material,  the  sec- 
tion i-i  will  fail.  It  is  usual  to  assume  that  the  resultant  R 
must  fall  within  the  distance  q  from  the  center  of  gravity  of  the 
section.  Had  the  force  F  been  applied  at  all  possible  points 
around  the  column  at  the  same  distance  y  above  the  section 
i-i  the  points  of  intersection  of  the  resultants  R  of  W  and  F 
would  have  described  the  figure  shown  black  in  Fig.  261  (b); 
this  is  called  the  kern  of  the  section.  If  the  section  i-i  at  no 
point  is  to  be  subjected  to  tension  the  resultant  R  must  fall 
within  the  kern  for  all  positions  of  the  force  F. 

The  moment  on  the  section  for  a  given  loading  will  be  the  load 
W  times  the  distance  from  the  center  of  gravity  of  the  section 
to  the  point  of  intersection  of  the  resultant  R  with  the  plane  of 
the  section  i-i. 

The  allowable  distance  q  from  the  center  of  gravity  of  the 
section  to  the  resultant  R  that  produces  zero  stress  at  the  extreme 

fibers  of  the  section  for  that  position  of  F  is  given  by  q  =  -      —  > 

A  X  c 

where  /  =  moment  of  inertia  of  the  section,  inches3. 
A  =  area  of  the  section,  square  inches. 
e  =  distance  from  center  of  gravity  of  section  to  extreme 
fibers  whose  stress  is  to  be  limited  to  zero. 

The  kerns  for  the  commonest  sections  used  for  chimneys  and 
foundations  are  given  in  Figs.  262  to  267. 

The  minimum  radii  for  the  kerns  of  these  several  sections  are, 

Square,  Fig.  262,  r  =  0.118  h. 

b-h 
Rectangle,  Fig.  263,  r  =  ' 


240  GRAPHICS  AND  STRUCTURAL  DESIGN 

Triangle,  Fig.  267,  r2  =  —  ,   and  n  =  -• 

12  6 

Octagon,  Fig.  264,  r^  =  0.2256  #  (R  =  radius  of  corners). 

Circle,  Fig.  265,  r  =-  (constant). 
8 

Hollow  square,  outer  side  =  H,  inner  side  =  h.     Similar  to* 
Fig.  262 


Circular  ring,  Fig.  266,  outer  diameter  =  D,  inner  diameter 


In  the  case  of  retaining  walls  it  is  necessary  to  consider  the  force 
F  as  acting  in  one  direction;  the  base  is  then  rectangular  and 
the  resultant  R  must  not  fall  farther  from  the  center  of  the  base 

than  one-sixth  the  base  width,  or  if  b  =  width  of  the  base,  q  =  - . 

o 

This  is  commonly  expressed  by  saying  that  the  resultant  pressure 
must  pass  within  the  middle  third. 

In  chimney  design  in  this  country  the  wind  pressure  is  gen- 
erally assumed  at  30  to  50  Ibs.  per  sq.  ft.  upon  flat  surfaces,  and 
at  20  to  30  Ibs.  per  sq.  ft.  on  the  projected  area  of  cylindrical 
surfaces.  The  greater  of  these  pressures  considerably  exceeds 
any  pressure  likely  to  occur  in  ordinary  localities. 

Ordinary  masonry  chimneys  are  commonly  either  square, 
octagonal  or  round.  Square  sections  are  suitable  for  short  and 
unimportant  chimneys  while  by  far  the  greater  number  of 
chimneys  are  round. 

Chimney  sections  are  obtained  by  the  use  of  specially  shaped 
brick  for  the  corners  of  octagonal  chimneys  and  radial  brick  for 
round  chimneys. 

In  small  chimneys  the  upper  25  ft.  should  have  a  thickness  of 
8  to  9  ins. ;  this  thickness  should  be  increased  about  4  to  4  J  ins. 


L 


" 


FIG.  262. 


|*--R-^TR--^| 


6 >j 

FIG.  263. 


\*--d— >| 
FIG.  264.  FIG.  265. 


FIG.  266. 


FIG.  267. 


- 


I 

i 


,Top 


.F 

30800^ 


a 


,  Ground 


BRICK  CHIMNEY 
FIG.  268. 


(241) 


242  GRAPHICS  AND   STRUCTURAL  DESIGN 

in  each  succeeding  lower  25  ft.  In  large  chimneys  whose  diam- 
eters exceed  60  ins.,  the  thickness  of  the  upper  25  ft.  should  be 
12  to  13  ins. 

Ordinary  brick  dimensions  are,  width  4  to  4!  ins. ;  ,  length 
8  to  9  ins. 

Chimneys  are  commonly  provided  with  a  fireproof  lining  in 
the  lower  portion  extending  up  to  from  one-third  to  one-half 
the  chimney  height.  This  lining,  being  protected  from  the  wind 
pressure  by  the  main  walls  of  the  chimney,  can  be  made  quite 
thin,  frequently  not  exceeding  from  4  to  8  ins. 

DESIGNING  A  CHIMNEY 

The  method  of  designing  a  chimney  will  be  illustrated  by 
making  the  calculations  for  the  section  d-d  of  the  chimney  in 
Fig.  268.  The  weight  of  the  material  above  the  section  has 
been  estimated  as  545,000  Ibs.,  being  taken  at  no  Ibs.  per  cu.  ft. 

The  wind  pressure  has  been  considered  as  25  Ibs.  per  sq.  ft. 
•of  projected  area  for  a  round  chimney  and  has  been  estimated 
as  30,800  Ibs.  This  wind  force  has  been  assumed  as  acting  at 
the  center  of  gravity  of  the  trapezoid  which  is  the  projected 
area  of  the  chimney  above  the  section. 

The  distance  from  the  section  d-d  to  the  center  of  gravity  is 

x  =  L+^l  X  d-  =   14.16  +  20  x  100  X  12  =  ms 

b  +  bi        3       14.16  +  10  3 

b   =  base  of  trapezoid. 

bi  =  top  of  trapezoid. 

d   =  altitude  of  trapezoid. 

From  the  intersection  of  the  chimney  axis  and  wind  force  lay 
off  hi  representing  the  weight  to  scale,  545,000  Ibs.,  and  at  its 
lower  extremity  lay  off  ij  to  the  same  scale  representing  F  = 
30,800  Ibs.  Draw  the  hypothenuse  of  this  right-angled  triangle 
and  it  will  cut  the  section  d-d  at  /.  To  insure  the  masonry  at 
this  section  not  being  subjected  to  tension  the  point  /  must  fall 
within  the  kern  of  this  section. 


CHIMNEYS 


243 


For  accuracy  it  is  better  to  calculate  the  length  Ik  than  to 
measure  it  from  a  drawing  on  so  small  a  scale.  The  calculation 
can  be  made  by  the  use  of  the  similar  triangles  hlk  and  hij,  hence 

„  hk  X  if      556  X  30,800 

Ik  =  q  =  — rr^1  =  -  -  =  31.4  ms. 

hi  545,000 

hk  =  distance  from  the  section  to  the  resultant  wind  force. 
The  kern  radius  is  given  by 


Since  31.4  ins.  is  less  than  35.6  ins.  it  is  evident  that  the 
resultant  falls  within  the  kern,  and  the  section  is  subjected  to 
compression  only. 

The  weights  have  been  calculated,  using  the  volume  of  a 
frustum  of  a  cone  as 

F  =  ^U-«H 


h  =  height  of  the  frustum. 
A  =  area  of  the  base. 
a  =  area  of  the  top. 

The  following  table  gives  the  properties  of  the  several  sections. 


Section  above. 

Distance 
from  top, 
ft. 

Diameter 
O.D. 
I.D. 

Mean 
diameter. 

Mean 
area, 
sq.  ft. 

Weight 
above,  Ibs. 

Fat 
25  Ibs. 

r" 

q" 

At  top  

0 

\  's'^' 

a-a  

25'  o" 

II'  4' 
9'  4' 

10'  8" 

8'  8" 

28 

77,ooo 

6,700 

28.6 

12.7 

b-b  

So'  o" 

12'  8' 
10'  o' 

12'  o" 
9'  8" 

40 

187,000 

13,750 

30.8 

21.3 

c-c 

75'  o" 

14'  o' 

13'  4" 

56 

341,000 

21,800 

33.2 

27.3 

d-d 

100'  o" 

10'  8' 
IS'  4' 
11'  4' 

10'  4  ' 
14'  8  ' 

II'   O  ' 

74 

545,000 

30,800 

35-6 

31-4 

e-e  

125'  o" 

16'  8' 
12'  o' 

16'  o  ' 
n'  8  ' 

94 

803,000 

40,700 

38.o 

34.8 

/-/  

150'  o" 

18'  o' 

12'   8' 

17'  4  ' 
12'  4  ' 

117 

1,127,000 

52,500 

40.1 

37-3 

(Max.1 

Top  of  foundation  .  .  . 

185'  o" 

18'  o" 
n'  o" 

229 

2,008,650 

67,500 

93.5, 
IMin.f 

33-4 

[66.  oj 

f    Top 

Base  foundation. 
Concrete  at  140  Ibs.  per  cu.  ft.. 

1  22'  o", 
1  Bottom 



2,867,130 

67,500 

(Min.j 

23-9 

I  3o'  o" 

244  GRAPHICS  AND   STRUCTURAL  DESIGN 

It  is  seen  in  the  above  table  that  since  the  resultant  pressure  al- 
ways falls  within  the  kern  the  maximum  pressure  never  exceeds 
twice  the  fiber  stress  due  to  the  weight  above  the  section  assumed 
as  uniformly  distributed  over  that  section.  Considering  the  sec- 
tion at/-/,  the  load  above  that  section  being  1,127,000  Ibs.,  and 
the  area  of  the  section  117  sq.  ft.,  the  load  per  square  foot  is 

1,127,000        ,      „ 
- — " =  9630  Ibs. 

This  is  well  below  the  allowable,  or  144  X  150  =  21,600  Ibs. 
per  sq.  ft. 

Some  designers  prefer  to  use  the  inertia  of  the  section  in 
making  their  calculations.  Making  the  calculations  for  the 
section  150  ft.  from  the  top  we  have 

The  section  modulus  of  the  ring 

=  /  _  TT  (£>4  -  <J4)  =  TT  (D2  -  d2)  (D2  +  d2) 
~  e~         32J9  4X8Z) 

Since  -  (D2  —  d2)  is  the  area  of  the  ring  section  which  we  can 
4 

call  A  we  have 


The  dimensions  for  the  section  150  ft.  from  the  top  are  A  =  117 
sq.  ft.,  D  =  1 8  ft.  o  ins.,  d  =  12.67  ft->  hence 


The  projected  area  of  the  chimney  is  -          -  X  150  =  2100  sq. 

ft.     The  wind  pressure  on  this  area  is  2100  X  25  =  52,500  Ibs. 
The  distance  x  from  the  section  to  the  center  of  gravity  is 


CHIMNEYS  245 

The  moment  of  the  wind  about  the  section  is  52,500  X  68  = 
3,570,000  ft.  Ibs.,  since 

f      Me      3.570,000  f. 

f  =  -  -  =  9100  Ibs.  per  sq.  ft. 

I  393 

The  direct  pressure  upon  this  section  is  —  -  =  9630  Ibs. 

per  sq.  ft.  The  maximum  pressure  upon  the  leeward  side  is 
9630  +  9100  =  18,730  Ibs.,  while  that  on  the  windward  side  is 
9630  —  9100  =  530  Ibs.  per  sq.  ft.,  both  being  in  compression. 

Some  designers  assume  an  allowable  tension  on  the  windward 
side  of  one-tenth  the  maximum  compression  on  the  leeward  side. 
It  is  then  understood  that  the  material  does  not  exert  this  tensile 
fiber  stress  but  that  the  effect  is  to  make  the  resultant  com- 
pression pass  outside  of  the  kern  and  increase  the  maximum  unit 
pressure.  This  can  be  understood  by  referring  to  Figs.  261  (a) 
to  261  (f). 

The  determination  of  the  maximum  pressure  under  these 
circumstances  presents  considerable  difficulties.  As  long  as  the 
resultant  pressure  at  any  section  fell  within  that  section  the 
margin  of  security  would  be  given  by  the  ratio  of  the  ultimate 
crushing  strength  of  the  material  to  the  extreme  fiber  stress 
in  compression  at  that  section.  Keeping  the  resultant  within 
the  kern  affords  a  ready  means  of  determining  the  maximum 
fiber  stress,  thus  assuring  safety. 

The  maximum  pressure  upon  the  foundation  will  now  be 
determined.  The  foundation  being  30  ft.  o  ins.  square  the  min- 
imum radius  of  the  kern  is 

r  =  0.118  X  30  X  12  =  42.5  ins. 

As  the  resultant  passes  23.9  ins.  from  the  center  line  of  the 
foundation  it  falls  in  the  kern  and  the  maximum  unit  compression 
will  not  exceed  twice  the  unit  uniform  load  on  the  foundation 
when  no  wind  is  blowing. 

The  maximum  unit  pressure  will  be  on  a  corner  when  the  wind 
blows  parallel  to  a  diagonal.  The  moment  on  the  foundation 


246 


GRAPHICS  AND   STRUCTURAL  DESIGN 


FIG.  269. 


-0.60- 


-0.40- 


-0.40 


-0.30- 


-0.30 


:0.20- 


J-L 


-0.20 


—0.15 


=.  0.2  0.3  0.4 

i  i  I  i  i  i  i  i  i  i  i  i  I  i  i i  I  i  i  i  i  i  i  i  i  i  I  i  i  i  i  l  i 


FIG.  270. 


CHIMNEYS 


247 


section  is  the  product  of  the  total  weight  on  the  soil  multiplied 

by  q  or 

1/r      2,867.000  X  23.0  .    „ 

M  =  -  -  =  5,720,000  ft.  Ibs. 


12 


The  resistance  of  a  square  referred  to  a  diagonal  as  an  axis  is 


-  =  0.118  h\ 
e 


The  base  being  30  ft.  square  and  keeping  the  resistance  in 
feet3  we  have, 


-  =  0.118  X  3o 
e 


3186. 


0.          ,      M-e      5,720,000  r,     ,,     j. 

Since  /  =  =  -   — =  1800  Ibs.  per  sq.  ft.,  the  direct 

1  3*"^ 

3190   Ibs.   per  sq.    ft.     The   maximum 

rh 


.      2,867,000 

pressure  is  - 

900 

compression  then  is  1800  -f-  3190  =  4990  Ibs. 
per  sq.  ft.  The  compression  on  the  opposite 
corner  is  3190  —  1800  =  1390  Ibs.  per  sq.  ft. 

These  loads  will  be  satisfactory  if  the  soil  can 
carry  2\  tons  per  sq.  ft. 

When  the  resultant  of  the  wind  pressure  and 
the  chimney  weight  passes  outside  of  the  kern 
of  the  bottom  of  a  chimney  the  maximum  pres- 
sure on  the  corner  of  a  square  or  the  circum- 
ference of  a  circular  section  may  be  found  by 
using  the  curves  given  in  Figs.  269  and  270. 

Fig.  269  represents  a  square  base,  Fig.  270  a 
circular  base. 

D  =  diagonal  of  square  and  diameter  of  cir- 
cle. 
h  =  side  of  square. 


FIG.  271. 


q  =  ph  =  (3D  =  distance  from  center  of  section  to  resultant 

of  wind  and  weight. 
a-a  —  neutral  axis,  axis  of  zero  pressure. 


248  GRAPHICS  AND  STRUCTURAL  DESIGN 

F  =  total  wind  force  on  chimney,  acting  at  the  center  of 
gravity  of  the  projected  area  of  the  exposed  portion  of 
the  chimney. 

W  =  total  load  on  the  soil,  pounds. 

hi  =  distance  from  force  F  to  bottom  of  the  foundation, 

7T< 

inches.     See  Fig.  271.     q  (inches)  =  hi  X  •=-. r- 

W 

SELF-SUSTAINING  STEEL  CHIMNEYS 

A  self-sustaining  steel  chimney,  Fig.  272,  is  held  upright  by 
the  resistance  to  overturning  offered  by  the  chimney,  its  lining 
and  foundation. 

The  wind  pressure  is  commonly  assumed  as  50  Ibs.  per  sq.  ft., 
acting  upon  the  side  of  a  square  chimney,  or  25  to  30  Ibs.  per 
sq.  ft.  of  vertical  projection  upon  round  chimneys. 

Usually  the  thickness  of  the  upper  sheets  is  determined  for 
durability  rather  than  for  strength,  being  made  not  less  than 
T3g  in.  and  frequently  \  in.  thick.  This  thickness  is  then  in- 
creased by  -IQ  in.  every  30  or  40  ft.  Some  designers  vary  by 
•£$  in.  instead  of  by  yg-  in.  At  the  lower  part  of  each  30  or  40  ft. 
section  the  fiber  stress  induced  in  it  by  the  bending  moment 
due  to  the  wind  pressure  can  be  found  and  the  section  altered  in 
thickness  or  length  if  deemed  advisable. 

The  rivet  spacing  is  made  small  to  assure  tightness,  being  not 
less  than  2.5  times  the  rivet  diameter  nor  more  than  16  times 
the  thickness  of  the  plate.  For  most  ring  sections  this  gives 
excessive  rivet  strength. 

An  assumed  ring  section  may  be  checked  as  follows: 

M  =  bending  moment  in  inch  pounds  on  the  section  due  to  the 
assumed  wind  pressure  above  that  section. 

D  =  outside  diameter  of  the  chimney  in  inches. 

h  =  distance  from  a  ring  section  to  the  chimney  top,  inches. 
/  =  thickness  of  shell,  inches. 

F  =  total  wind  pressure  acting  upon  any  portion  of  the 
chimney,  pounds. 


FIG.  272. 


- H 

SECTION  AT  a-a 

FIG.  274. 


BLRA.CKET 

STEEL  CHIMNEY 


(240) 


250  GRAPHICS  AND   STRUCTURAL  DESIGN 

The  chimney  above  the  given  ring  section  acts  as  a  cantilever 
beam,  and  the  bending  moment  equals  the  total  wind  pressure 
upon  the  vertical  projection  of  the  chimney  above  this  ring 
section  multiplied  by  one-half  the  distance  from  the  ring  section 
to  the  chimney  top.  The  chimney  will  be  assumed  66  ins. 
O.D.  and  100  ft.  high.  The  bell  section  is  9  ft.  high  and  the 
section  to  be  examined  is  91  ft.  from  the  chimney  top.  The 
total  wind  pressure  on  the  vertical  projection  of  this  part  of 
the  chimney  is  F  =  5.5  X  91  X  25  =  12,510  Ibs.  and 

M  =  12,510  X  -          -  =  6,830,460  in.  Ibs. 

The  value  of  -  for  a  ring  section  is 
e 

I      TT  CD4  -  d*) 


^ 
e         32  XD 

where  D  =  outside  diameter. 
d  =  inside  diameter. 

Where  /,  the  thickness  of  the  shell,  is  small  compared  with  D, 
this  value  may  be  approximated  as 


-  for  the  case  in  hand  is  -  =  -  X  66.625  X  -^  (66.625  -  0.9375) 
e  e      5  16 

=  1094. 

f      Me      6,8^0,460      , 
/  =  --.=  -2-£-     -  =6250  Ibs. 
/  1094 

This  fiber  stress  is  on  the  gross  section.  If  the  net  section  at  the 
rivet  circle  is  x  per  cent  of  the  total  section  then  the  maximum 
fiber  stress  between  rivets  is 

f       6250  ..  ,        6250 

fm  =  -   -,    or,  if    x  =  80  per  cent,    fm=—^- 

X  .oO 

=  7800  Ibs.  per  sq.  in. 

As  these  fiber  stresses  are  satisfactory  the  y\-in.  shell  will  be 
used. 


CHIMNEYS  251 

RIVETING  RING  SEAMS 

5  =  pitch  of  rivets,  inches.  The  force  acting  on  the  rivets 
per  inch  of  circumference  is  /  X  /  X  /,  and  to  this  must  be  added 
the  weight  of  the  shell  i  in.  wide  from  the  given  section  to  the 
top.  The  calculation  of  this  weight  can  be  facilitated  by  remem- 
bering that  a  J-in.  plate  12  ins.  X  12  ins.  weighs  10.2  Ibs.,  then 
estimating  the  weight  of  a  vertical  strip  12  ins.  wide  and  finally 
dividing  the  weight  found  by  12. 

For  the  section  91  ft.  from  the  top,  /  X  /  X  p  =  /  X  fV  X  6250 
=  1950  Ibs. 

1.  30  ft.  of  TVm-  plate,  30  X  i  X  (J  X  10.2)  =      230 

2.  30  ft.  of    J-in.  plate,  30  X  i  X  (i  X  10.2)  =      306 

3.  30  ft.  of  j^-in.  plate,  30  X  i  X  (|  X  10.2)  =      383 

Total  919 

Adding  10  per  cent  for  rivets,  laps,  etc.  91 


1010 

or  -  -  =  84  Ibs.  per  in.  The  total  force  per  inch  of  circum- 
ference acting  on  the  rivets  is  1950  -f-  80  =  2030  Ibs.  Allowing 
10,000  Ibs.  per  sq.  in.  in  single  shear  and  20,000  Ibs.  per  sq.  in. 
in  bearing  the  value  of  a  f-in.  rivet  in  a  -j^-in.  plate  is  4420  Ibs., 

and  the  rivet  spacing  required  is  -     -  =  2.18  ins.  or,  say,  not 

2030 

exceeding  i\  ins. 

A  double  row  of  staggered  rivets  will  be  used  at  this  seam 
spaced  about,  but  not  exceeding,  2\  ins.  Ordinarily  the  rivet 
spacing  need  be  determined  for  only  the  lowest  row  of  rivets  in 
sheets  of  the  same  thickness. 

RIVETING  VERTICAL  SEAMS 

The  riveting  in  the  vertical  seams  must  provide  for  the  shear 
along  the  neutral  axis  of  the  chimney,  and  as  the  riveted  edge 
may  be  in  compression  the  rivets  must  be  spaced  to  prevent 


252  GRAPHICS  AND   STRUCTURAL  DESIGN 

the  buckling  of  the  plate  between  them;  this  demands  that  the 

rivet  spacing  shall  not  exceed,  1  6  times  the  thickness  of  the  plate. 

The  unit  shearing  stress  at  any  point  along  a  cylindrical  beam 

V  -r 

~T 

e 

where  /,  =  unit  shear  in  pounds  per  square  inch  at  the  section. 
V  =  shear  at  right  angles  to  the  neutral  axis,  pounds. 
r  =  radius  at  the  section,  inches. 

-  =  resistance  of  the  section,  inches3. 
e 

In  this  chimney  at  the  section  under  consideration  the  unit 
shearing  stress  is 

=    8olbs. 


1094 


This  per  inch  of  chimney  height  is  380  X  YG  =  1  2O  Mbs. 

The  rivet  spacing  for  shear  then  is  —  —  ~  37  ins. 

1  20 

This  spacing  of  course  is  not  permissible,  16  times  the  plate 
thickness  limiting  the  spacing  to  5  ins.  It  generally  will  be  the 
buckling  of  the  plate  edge  that  will  determine  this  riveting. 

The  ring  sheets  are  preferably  of  one  piece,  but  in  large  chim- 
neys this  becomes  impractical  both  for  manufacture  and  ship- 
ment. 

The  base  of  the  chimney  is  usually  a  bell,  and  is  preferably 
a  frustum  of  a  cone,  although  sometimes  flared  to  improve  the 
appearance.  The  former  shape  is  a  much  simpler  and  cheaper 
design.  The  height  of  the  bell  and  the  diameter  at  its  base 
vary  from  ij  to  2  times  the  chimney  diameter.  The  bell  sec- 
tion is  usually  made  of  a  number  of  pieces,  butt  jointed  with 
single  outside  straps  as  shown  in  the  drawing.  The  bell  sheets 
are  usually  tV~m-  thicker  than  the  ring  sheets  immediately  above 
them. 


CHIMNEYS 


253 


Where  the  tempera- 


T 


Lining.  —  The  method  of  lining  varies, 
tures  do  not  exceed  650°  F.  (and  they 
ordinarily  do  not)  common  red  brick 
are  satisfactory;  otherwise  a  No.  2 
fire  brick  is  required.  At  the  top  the 
lining  should  have  a  thickness  4!  ins., 
which  .can  be  increased  by  4^  ins. 
every  30  to  40  ft.  In  some  cases  the 
chimney  is  only  partially  lined,  say 
one-half  the  way  up,  and  in  other 
cases  the  thinnest  lining  is  made  2\ 
ins.;  the  former,  however,  is  the  bet- 
ter practice. 

Where  the  breeching  from  the  boil- 
ers enters  the  chimney  above  the  base 
care  should  be  taken  not  to  weaken 
the  chimney  by  cutting  away  too 
much,  and  the  chimney  should  be 
sufficiently  reinforced  at  this  point. 
The  practice  of  having  the  gases  enter 
the  chimney  through  flues  below  the 
steel  base  makes  the  chimney  stronger 
and  of  better  appearance. 

FOUNDATIONS 

The  shell  is  maintained  upright 
against  the  moment  of  the  wind  pres- 
sure by  the  weight  of  the  foundation 
and  chimney  with  lining  assumed  as 
acting  about  the  lower  edge  of  the 
foundation. 

The  foundation  is  concrete  and  the 
chimney  is  secured  to  it  by  heavy 
foundation  bolts.  The  concrete  can 
be  assumed  as  weighing  from  125  to  150  Ibs.  per  cu.  ft.  When 


Fig.  277  (d) 


254  GRAPHICS  AND   STRUCTURAL  DESIGN 

there  is  no  wind  acting  upon  the  chimney  the  combined  weights 
of  the  foundation,  Wf,  the  steel  chimney,  Wc,  and  the  lining, 
Wi,  are  distributed  uniformly  over  the  soil  through  the  base  of 
the  foundation,  as  is  shown  in  Fig.  277  (b).  Now  as  a  wind 
pressure  F  begins  to  act  the  uniform  pressure  previously  on  the 
soil  will  be  altered,  the  pressure  being  reduced  on  the  side  upon 
which  the  wind  acts  and  increased  on  the  opposite  side.  In 
Fig.  277  (b)  5  and  Wf  +  Wc  +  Wh  which  are  always  equal,  lie 
in  the  same  line.  In  Fig.  277  (c)  the  force  acting  upon  the 
base  from  the  soil,  S.  nas  moved  over,  acting  through  the  center 
of  gravity  of  the  trapezoid  which  here  represents  the  distribu- 
tion of  the  soil  pressure. 

When  the  wind  pressure  becomes  great  enough  the  distribu- 
tion of  pressure  on  the  soil  will  become  that  shown  in  Fig.  277  (d), 
being  zero  at  the  right,  while  the  original  pressure  at  the  left  is 
doubled.  The  force  S  will  pass  through  the  center  of  gravity 
of  the  triangle,  or  J  B  from  the  edge  of  the  foundation.  It  is 
commonly  assumed  in  masonry  construction  that  the  resultant 
pressure  upon  any  section  must  pass  within  the  kern  of  that 
section.  Were  the  resultant  S  to  pass  to  the  left  it  would  pass 
out  of  the  kern.  See  page  238.  In  Fig.  277  (a)  the  moment 
due  to  the  wind  force  must  equal  the  moment  of  the  forces  Wf, 
Wc  and  Wi,  with  the  arm  a;  hence 


(Wf  +  Wc  +  Wl)  a; 

the  limit  being  a  =  0.118  B  we  have 

F(^  +  Hf\  =  (Wf  +  Wc  +  Wl}  Xo.nSB. 

The  volume  of  a  frustum  of  a  cone  or  pyramid  is 


V  =  (At  +  Ab 

o 

where         At  =  area  of  the  top  of  the  frustum. 

Ab  =  area  of  the  bottom  of  the  frustum. 
h    =  altitude  of  the  frustum. 


CHIMNEYS  255 

Maximum  Pressure  on  the  Soil.  —  When  the  resultant  pressure 
passes  through  the  kern  the  maximum  pressure  will  be  less  than 
twice  the  mean  pressure.  The  maximum  pressure  per  square 
foot  then  is 


P    =  pounds  per  square  foot. 

Ab  =  area  of  the  base  of  foundation,  square  feet. 

Where  the  foundation  is  not  reinforced  the  angle  of  the  sides  with 
the  vertical  should  not  exceed  30  degrees. 

The  maximum  pressure  should  not  exceed  that  permitted  on 
the  soil  (see  page  228). 

In  the  chimney  under  discussion  assume  the  base  18  ft.  square 
and  that  the  concrete  weighs  140  Ibs.  per  cu.  ft.  Estimating  the 
weight  of  the  steel  of  the  chimney,  the  diameter  being  5  ft.  6  ins., 
the  circumference  is  17.3  ft. 

Sheets.  Dimensions,  Weights, 

feet.  Ibs. 

T8<5  17-3  X  30  X  10.2  X  |  3,980 

i  17-3  X  3°  X  10.2  X  i  5,300 

rV  J7-3  X  3°  X  10.2  X  £  6,620 

f  17.3  X  10  X  10.2  X  £  2,650 


18,550  Ibs. 
Allowing  10  per  cent  for  bolts,  rivets,  laps,  etc.,  1,850 

Total  20,400  Ibs. 

The  weight  of  the  lining  assuming  the  chimney  to  be  lined  to 
a  height  of  70  ft., 

Section.                           Lining,  Area  of  lining.  Volume, 

ins.                      sq.  ft.  cu.  ft. 

Base  to  40  ft.                9                      11.2  448 

40  to  70  ft.                    4^                      6.05  135 

Total  583 

Weight  of  lining,  580  X  125  =  72,500  Ibs. 

The  height  of  the  foundation  will  be  assumed  at  10  ft. 


256  GRAPHICS  AND   STRUCTURAL  DESIGN 

If  the  resultant  of  wind  pressure  and  total  weight  is  to  fall 
within  the  kern  its  distance  from  the  center  line  of  the  chimney 
must  not  exceed  0.118  X  B  =  0.118  X  18  X  12  =  25.5  ins.; 
this  is  the  distance  along  the  diagonal.  At  right  angles  to  the 

h  12 

side  of  the  square  it  would  be-=i8X  —  =36  ins. 

6  6 

Trying  the  following  foundation,  top  10  ft.  square,  bottom 
1  8  ft.  square  and  height  10  ft.,  the  volume  of  this  foundation 
is 

V  =  -  (At  +  Ab  +  ^AtXAb  =  —  (io2  +  i82  +  Vio2  X  i82) 

o  o 

=  2013  cu.  ft. 
Weight  =  2013  X  140  =  281,800  Ibs. 

The    total   weight   on    the   soil   is    281,800  +  20,400  +  72,500 

=  374,700  Ibs.     The  uniform  pressure  per  square  foot  is  374>7°o 

324 

=  1150  Ibs.  The  resultant  of  wind  pressure  and  total  weight 
cuts  the  bottom  of  the  foundation  a  distance  q  from  the  center 
line  of  the  chimney.  The  total  wind  force  is  F  =  5.5  X  100  X  30 
=  16,500  Ibs.  This  force  acts  at  a  distance  of  50  ft.  from  the 
base  of  the  chimney  or  60  ft.  from  the  bottom  of  the  foundation. 

TT  60  X  12  X  16,500 

Hence  q  =  —  -  L2—  =  31.7  ins. 

374,700 

The  resultant  evidently  falls  outside  the  kern  on  the  diagonal 
but  passes  within  it  at  right  angles  to  the  side. 

The  maximum  pressure  on  the  corner  can  be  found  by  the 
assistance  of  the  curves,  Fig.  269. 


From  the  curves  for  a  square  base  when  ft  =  0.147,  «  =  °-22> 
and  since 

-  Per  sq.ft. 


CHIMNEYS 


257 


the  foundation  will  be  satisfactory  if  this  load  of  2625  Ibs.  per 
sq.  ft.  is  permissible  upon  the  soil  under  the  chimney. 

FOUNDATION  BOLTS 

The  chimney  is  assumed  as  tending  to  overturn  about  the 
axis  a-a  tangent  to  the  bolt  circle  and  differing  only  slightly 
from  the  lower  edge  of  the  bell. 
The  maximum  fiber  stress  will 
be  fm  and  will  occur  in  bolts 
No.  3  and  No.  4  (see  Fig.  278), 
these  being  farthest  from  a-a. 
The  fiber  stress  in  any  other    / 


r- 


FIG.  278. 


bolt  will  be  fm  multiplied  by 
the  ratio  of  its  distance  from 
a-a  to  the  distance  of  the  bolts 
farthest  from  a-a]  thus  the 
fiber  stress  in  the  bolts  2  and 

5  is/m  X  -.     The  resisting  mo- 

p 

ment  due  to  any  bolt  is  the  product  of  its  area,  the  fiber  stress 
acting  in  it  and  its  distance  from  the  axis  a-a.  The  resisting 
moment  of  the  entire  group  of  bolts  will  be  the  sum  of  the 
moments  of  the  individual  bolts.  Taking  the  case  of  the  six 
bolts  shown,  and  letting  a  be  the  area  of  one  bolt,  at  the  root 
of  the  thread,  and  R  the  radius  of  the  lower  bell  circle,  we  have, 

No.  bolts.  p  Fiber  stress.  Moment. 

i  and  6     R  (i  -  cos   .,0°)  =  0.134  R       fmX  ^f|4 

1.866  K 

R 


R  (i  —  cos  30°)  =  0.134  R 
2  and  5     R  (i  —  cos  90°)  =  R 


3  and  4     R  (i  -  cos  150°)  =  1.866 R      fmXi 


2-a-fm-R  ^~ 
2-a-fm-R-i.S66 


This  gives  a  total  moment  of  M  =  4.8  X  a  *fm  •  R. 

In  this  way  it  may  be  shown  that  in  the  expression  M  — 
C  •  a  *fm  •  R,  C  has  the  following  values  for  different  numbers 
of  bolts. 


c. 

Number  of  bolts. 

C. 

3-5 
4.8 

12 

16 

9.1 

12.0 

6.2 

20 

15-0 

7-7 

258  GRAPHICS  AND   STRUCTURAL  DESIGN 

Number  of  bolts. 

4 

6 

8 

10 

The  moment  on  the  chimney  due  to  the  wind  has  been  found  to 
be  6,830,460  in.  Ibs.  The  moment  due  to  the  weight  of  the* 

chimney  which  tends  to  hold  it  upright  is  MI  =  20,400  X  — 

2 

=  918,000  in.  Ibs.  The  net  moment  to  be  resisted  by  the  bolts 
is  6,830,460  —  918,000  =  5,912,460  in.  Ibs. 

Trying  four  bolts,  M  =  4.8  a  .fmR,  or  a  =  5'9I2>46° = 

4,8  X  9000  X  47 

2.91  sq.  ins.  This  corresponds  to  2j-in.  diameter  bolts.  The 
maximum  total  tension  in  a  single  bolt  is  2.91  X  9000  = 
26,190  Ibs. 

The  bracket  through  which  the  bolt  fastens  to  the  bell  of  the 
chimney  must  be  designed  to  resist  this  force.  A  very  effective 
design  is  shown  in  Fig.  276.  The  bolts  should  be  kept  back  as 
close  as  possible  to  the  edge  of  the  bell,  and  the  bell  sheet  amply 
reinforced  with  straps  at  this  point.  The  lower  edge  of  the  bell 
should  be  reinforced  with  a  circular  band.  This  lower  edge 
then  rests  upon  a  cast-iron  base  plate,  Fig.  273,  that  distributes 
the  pressure  over  the  top  of  the  foundation.  The  base  plate  may 
be  made  in  one  piece  for  small  chimneys,  as  shown,  but  is  usually 
made  of  a  number  of  sections  for  convenience  in  casting -and 
handling.  These  sections  when  placed  in  position  are  secured 
with  a  few  J-in.  diameter  bolts.  After  erection  the  bell  of  the 
chimney  and  the  foundation  bolts  will  hold  these  sections  in 
place. 

The  foundation  bolts  should  extend  almost  to  the  bottom  of 
the  foundation  and  should  be  fitted  with  a  lock  nut  and  washer 
or  key  and  washer. 

The  chimney  should  have  a  light  ladder  running  to  a  platform 
at  the  chimney  top.  Dropping  the  platform  a  little  below  the 
top  protects  the  handrail  from  the  chimney  gases. 


CHIMNEYS  259 

REINFORCED-CONCRETE  CHIMNEYS 

In  a  reinforced-concrete  chimney  the  weight  of  the  concrete 
above  the  section  under  consideration  will  produce  a  compression 
on  that  section  similar  to  that  produced  in  a  masonry  chimney. 
Take  any  unit  area  having  a  ratio  of  reinforcement  of  0;  under  a 
given  load  the  shortening  of  the  two  materials  will  be  equal  and 
the  loads  shared  proportionally  to  their  moduli  of  elasticity. 

Let/c  =  compressive  unit  stress  in  concrete,  pounds  per  square 

inch. 
j? 

— -  =  n;  then  the  fiber  stress  on  the  steel  is  /„  =  n  •/.  and 
£c 

if  L  is  the  load  on  i  sq.  in.  of  section  fc  =  —  — -.     The 

i  +  <f>  (n  -  i) 

load  can  be  assumed  as  the  weight  of  the  concrete  for  the  ordi- 
nary percentages  of  steel  reinforcement. 

With  wind  acting  on  one  side  of  a  chimney  the  pressure  upon 
the  leeward  side  is  increased  while  that  on  the  windward  side 
is  decreased,  finally  becoming  zero  while  that  upon  the  leeward 
side  becomes  double  the  original  direct  unit  pressure.  The 
concrete  not  being  assumed  as  resisting  tension  may  however 
be  considered  as  acting  like  a  beam  until  the  compression  at  the 
windward  side  becomes  zero.  Under  these  circumstances  the 
resistance  of  the  section  may  be  used  as  in  the  case  of  a  beam 
resisting  both  tension  and  compression.  In  calculating  the 
moment  of  inertia  of  the  section  proper  allowance  must  be  made 
for  the  modulus  of  elasticity  of  the  steel  differing  from  that  of 
the  concrete. 

/  =  Ie  +  /.  =  ^  (D*  +  d*)  +  0.394  n  •  A2fc  (A  -3  0 ; 

here,  7  =  total  inertia  of  concrete  and  steel. 

7C  =  inertia  of  concrete  ring. 

7,  =  inertia  of  steel. 

A  =  area  of  concrete  ring,  square  inches. 
D  =  outside  diameter  of  section,  inches. 


260 


GRAPHICS  AND   STRUCTURAL  DESIGN 


d  =  inside  diameter  of  section,  inches. 
DI=  diameter  of  steel  cylinder  having  same  area  as  total 

reinforcement  at  the  section. 
t\  —  thickness  of  equivalent  cylindrical  steel  reinforcement. 

Generally  h  will  be  so  small  that  3^1  may  be  neglected,  making 
I  —  —  (D2  +  d2)  +  0.394  nD\t\. 

Where  the  ratio  of  reinforcement  is  known  t\  =  -(D  —  d). 

2 

When  the  point  has  been  reached  where  the  extreme  fiber 
stress  on  the  concrete  on  the  windward  side  is  zero  any  further 
increase  of  the  wind  pressure  F  will  be  resisted  by  increased 
compression  in  the  concrete  on  the  leeward  side  and  by  the  re- 
inforcing steel  taking  tension  on  the  windward  side.  This  causes 
the  axis  of  zero  stress  to  move  across  the  section  from  the  extreme 
fibers  on  the  windward  side  towards  the  leeward  side.  Consider- 
ing now  only  the  additional  moment  applied  after  the  extreme 
fiber  stress  on  the  windward  side  becomes  zero  there  will  be 
created  additional  compression  on  the  leeward  side  shown 
shaded,  while  on  the  windward  side  the  fiber  stress  will  be  tensile 
and  must  be  carried  by  the  steel.  Con- 
sidering now  the  fiber  stress  as  indicated 
in  Fig.  279,  it  will  vary  uniformly  on 
each  side  of  the  neutral  axis  a-a,  the 
stress  in  the  steel  being  n  times  the  stress 
in  the  concrete  at  the  same  distance  from 
the  axis  a-a. 

The  resultant  of  the  flange  stresses  in 
compression,  shown  in  the  shaded  por- 
tion in  Fig.  279,  must  equal  the  resultant 
of  the  tensile  stresses  in  the  steel  in  the 
remaining  part  of  the  section,  the  portion 

not  shaded.     The   resisting   moment   induced  in  any  section 
equals  either  flange  force  multiplied  by  the  distance  between 


CHIMNEYS  261 

the  lines  of  action  of  the  resultant  flange  forces  on  each  side  of 
the  axis  a-a.  This  distance  has  been  found  to  be  constant  and 
to  equal  1.56  R8,  where  R8  =  radius  of  the  steel  circle. 

The  values  of  the  flange  forces  and  the  moments  have  been 
calculated  based  upon  the  assumption  that  the  stresses  vary 
proportionally  to  the  distances  of  the  fibers  from  the  neutral 
axis  a-a,  and  that  the  fiber  stress  in  the  steel  is  n  times  that 
in  the  concrete  for  similar  positions,  the  thickness  of  the  wall 
RI  —  r  (Fig.  281)  has  been  considered  small  compared  with  R, 
that  is  the  flanges  have  been  considered  as  lines.  The  flange 
forces  are 

Tensile  steel,  Ft  =  2  t,ftRa  X  C\. 
Compression  steel,  Fc  =  2  tafcRa  X  C3. 
Compression  concrete,  Fco  =  2/cotcoR  X  C3. 

The  moments  due  to  these  forces  are 

Tensile  steel,   Mt  =  2/ttaRa2  X  C2. 
Compression  steel,  Mc  =  2fctaRa-  X  C4. 
Compression  concrete,  M^  =  2/cotcoR2  X  C4. 

Here  ta  =  thickness  of  equivalent  cylindrical  steel  reinforcement, 

inches. 

ft=  extreme  fiber  tension  in  steel,  pounds  per  square  inch. 
R8  =  radius  of  steel,  inches. 
fc  =  extreme  fiber  compression  in  steel,  pounds  per  square 

inch. 

tco  =  thickness  of  concrete,  inches. 
fco=  maximum  compression  in  concrete  at  mean  radius  R, 

pounds  per  square  inch. 
R  =  mean  radius  of  concrete,  inches. 

The  values  of  these  coefficients  Ci,  C^  Ca  and  C4  have  been  cal- 
culated for  the  several  values  of  k  and  are  given  in  the  curves, 
Fig.  280.  The  following  relations  are  important.  Ft  =  Fc  +  Fco. 
The  total  resisting  moment  of  a  section  M  —  Mt  +  Mc  +  M^. 


262 


GRAPHICS  AND   STRUCTURAL  DESIGN 


The  distance  between  the  lines  of  action  of  the  resultants  of  the 
flange  forces  has  been  previously  stated  as  1.56^;  hence  the 
moment  may  be  expressed  as 

(Fc  +  Fco)i.$6R. 


.80 


,20 


M=Fxl.5GR,s 


RE-INFORCED  CONCRETE 
CHIMNEYS 


Scale  on 


\ 


left 


multi  >lled  by  100 


.30 


07( 


Oil) 


FIG.  280. 

The  table  also  gives  the  relations  between  the  several  fiber 
stresses  and  the  critical  percentages  of  reinforcements  for  the 

various  values  of  k.     k  =  —.    Here  R  is  the  radius  of  the  flange 

R 

being  considered,  that  is,  Ra  for  the  steel  and  the  mean  radius 
for  the  concrete.  The  following  problem  will  illustrate  the  cal- 
culations of  a  chimney  as  outlined. 


CHIMNEYS 


263 


Problem.  —  A  chimney  is  1 50  ft.  high.  Its  outside  diameter 
is  10  ft.  and  the  walls  at  its  base  are  9  ins.  thick.  Assume  a 
wind  pressure  of  50  Ibs.  per  sq.  ft.,  giving  an  equivalent  pressure 
of  30  Ibs.  per  sq.  ft.  of  projected  area  of  the  round  chimney. 
The  pressure  on  the  concrete  is  not  to  exceed  500  Ibs.  per  sq.  in., 
while  the  unit  stress  on  the  steel  must  not  be  over  15,000  Ibs.  per 
sq.  in.  The  chimney  walls  are  6  ins.  thick  for  the  upper  100  ft. 
and  9  ins.  thick  for  the  lower  50  ft.  In  Fig.  282  assume  a  rein- 
forcement of  i  per  cent  and  that  a  bar  of  concrete  i  sq.  in.  in 


114—- 


FIG.  281. 


FIG.  282. 


section  and  12  ins.  high  weighs  i  Ib.     The  direct  compression  in 
the  concrete  at  the  base  of  the  chimney  is 

(100  X  |)  +  50       117 

pe  =  -        —^  -  ~  ioo  Ibs.  per  sq.  in. 

i  +4>(n  -  i)         1.14 

The  inertia  of  the  section  with  i  per  cent  steel  is 


/  = 


(i202  +  io22)  +  0.394  X  15  X  ii43  X  0.09 


13,350,000  in.  Ibs. 


,,      pi 

M  =  ^— 
e 


ioo  X  13.350.000  .      „ 

-  -  =  22,  200,000  in.  Ibs. 

60 


The  total  wind  force  is  W  =  10  X  150  X  30  =  45,000  Ibs.     The 


T  9 


moment  due  to  this  wind  pressure  is  M  =  45,000  X  150  X  — 

2 


264  GRAPHICS  AND  STRUCTURAL  DESIGN 

=  40,500,000  in.  Ibs.  The  remaining  moment  to  be  resisted  by 
the  reinforced  section  after  the  stress  in  the  extreme  fibers  on 
the  windward  side  has  become  zero  is  40,500,000  —  22,200,000 
=  18,300,000  in.  Ibs.  The  compression  on  the  leeward  side  at 
this  time  is  2  X  100  =  200  Ibs.  per  sq.  in.  The  permissible  in- 
crease in  the  compression  on  the  leeward  side  is  500  —  200  = 
300  Ibs.  per  sq.  in. 

The  additional  flange  force  is 

„       18,300,000      18,300,000 

F  =  (     \  ^  ^  =  ~  ~  =  2o6>000  lbs- 

(1.56X12)        1.56X57 

Ft  =  2  X  t.  Xfs  X  R8  X  Ci  =  2  X  0.09  X  15,000  X  57  X  1.32 

=  203,000  Ibs. 

This  is  sufficiently  near  the  required  206,000  Ibs.  From  the 
curves,  Fig.  280,  the  ratio  of  stress  in  the  steel  to  that  in  the 
concrete  is  57  to  i;  hence 

15,000 
pc  =  — —  =  263  Ibs.  per  sq.  in. 

The  approximate  total  stress  in  the  concrete  =  200  +  263 
=  463  Ibs.  It  should  be  noted  that  the  263  Ibs.  is  the  flexural 
stress  at  a  distance  R8  from  the  center  of  the  chimney  and  that 
this  can  be  easily  corrected. 

The  value  of  k  from  the  curves  corresponded  to  0.58,  making 

A  =  k  X  R3  =  0.58  X  57  =  33  ins. 

The  distance  from  the  neutral  axis  to  the  steel  57  —  33  =  24  ins. 
The  distance  to  the  outside  of  the  concrete  is  60  —  33  =  27  ins. 
By  similar  triangles, 

-%-x  =  ^,     or    /max  =  2961bs. 

fc  24 

The  extreme  fiber  stress  then  becomes  200  -f-  296  =  496  Ibs.  per 
sq.  in. 


CHIMNEYS  265 

DESIGN  OF  A  REINFORCED-CONCRETE  CHIMNEY 

In  Fig.  283,  assuming  the  chimney  wall  5  ins.  thick  at  the  top, 
the  area  of  this  section  is  1210  sq.  ins.     The  radius  of  the  kern 


This  means  that  before  any  reinforcement  would  be  required  the 
resultant  of  wind  force  and  weight  could  pass  18.3  ins.  beyond 
the  axis  of  the  chimney.  It  follows  that  the  weight  above  a 
section  /  feet  from  the  top,  divided  by  the  wind  force  acting  on 

the  portion  above  that  section  equals  -ins.,  divided  by  18.3,  or 
/ 

1210  X  /        =      2  .  =   1210  X  18.3  X  2  =      o        rt 

30X6.8  Xl      18.3'  30X6.8  X  12  . 

Hence  the  upper  18  ft.  would  require  no  reinforcement.  It  is 
usual  to  place  a  small  reinforcement  in  this  portion.  Assuming 
a  reinforcement  used  here  of  |  per  cent  (0.005)  the  distance  that 
this  will  serve  down  the  chimney  can  be  found  in  a  similar  way. 
From  the  curves  the  ratio  of  the  fiber  stresses  for  p  =  0.005  ^s 
85  to  i  ;  hence  taking  the  fiber  stress  in  the  steel  at  15,000  Ibs.  per 

sq.  in.  would  make  the  value  of  fe  =  I5'OOQ  =177  Ibs.  per  sq. 

85 

in.  ;  as  this  is  very  low  the  tensile  steel  will  determine  the  strength 
of  the  section.  The  value  of  Ci  is  1.38.  The  flange  force  in 
the  tensile  steel  then  is 

Ft  =2  tjt  X  R8  X  Ci  =  2  X  0.0025  X  15,000  X  38.5  X  1.38 
=  39,800  Ibs. 

Mi  =  Ft  X  1.56  X  R8  =  39,800  X  1.56  X  38.5  =  2,  395,  ooo  in.  Ibs. 
The  inertia  of  the  section  is 

/  =  A  (D*  +  d*)  +  0.394  n/Vd- 
.16 

+  7'2)  +  0<394  x  I5  x  7?a  x 


Fig.  283 


Fig.  284 


"I 


Fig.  285 


REINFORCED 
CONCRETE  CHIMNEY, 


CHIMNEYS  267 

Now  M  =  — ,  and  when  the  fiber  stress  on  the  windward  side 
e 

becomes  zero  that  on  the  leeward  side  has  become  double  the 
stress  due  to  the  weight  above  the  section. 

If  the  concrete  weighs  144  Ibs.  per  cu.  ft.,  and  the  distance 
from  the  top  to  the  section  is  /  feet,  then  /  =  /,  and  we  have 

//      /  X  968,406 

M  2  =  —  =  -  -  =  23,600 /. 

e  41 

The  bending  due  to  the  wind  acting  on  the  portion  above  the 
section  equals  the  sum  of  Mi  and  MZ.  From  the  wind  pressure 
and  the  chimney  dimensions  tht  bending  equals 


2 

Equating  these  values  of  M  we  have 

1220  X  I2  =  23,600  X  /  +  2,395,000, 
from  which 

I2  -  (19.3  X  /)  +  (^]   =  2040,     or    /  =  54.8  ft. 


This  reinforcement  will  have  an  area  of  1210  X  0.005  =  6.05  sq. 
ins.,  and  will  require  14  —  f-in.  round  bars.  Allowing  a  bond 
stress  of  80  Ibs.  per  sq.  in.  the  length  of  the  bars  at  laps  is 


where  li  =  length  of  the  lap  in  inches. 

/.  =  fiber  stress  in  steel,  pounds  per  square  inch. 
fb  =  bond  stress,  pounds  per  square  inch. 
d  =  diameter  of  round  bar  or  side  of  square  bar,  inches. 

In  the  problem 

,    _  12,000  X  d  _  , 

d  being  J  in.  the  lap  should  be  37.3  X  0.75  =  28.1  ins. 


268  GRAPHICS  AND  STRUCTURAL  DESIGN 

The  reinforcement  will  now  be  determined  for  the  section  at  the 
base.  The  moment  due  to  wind  at  this  section  of  the  chimney 
is 

M  =  [(85  X  6.8  X  30)  (^  +  40)]  +  [(40  X  8.5  X  30  X  20)] 

=  1,634,550  ft.  Ibs. 
M  —  19,570,600  in.  Ibs. 

D  =  102  ins.  d  =  88  ins.  The  area  of  the  section  is  2085  sq. 
ins.  The  weight  of  the  upper  85  ft.  is  1210  X  85  =  103,000  Ibs. 

The  load  per  square  inch  on  the  lower  section  is/c  =  I03>000  _j_  4O 

2085 

=  50  +  40  =  90  Ibs.  per  sq.  in.  Trying  a  reinforcement  of 
0.0175,  from  the  curves  C\  =  1.25,  j-  '=  41,  and  k  =  0.46. 

fa 

Revising  fc  for  the  steel  percentage 

.  _  90  _  90  _ 

Jc  ~  i. +  00-  i)  ~  i  +  (0.0175  x  14)  ~  72< 

The  compression  on  the  leeward  side  when  the  tension  on  the 
windward  side  is  zero  is  2  X  72  =  144  Ibs.  per  sq.  in.  The 
tension  in  the  steel  corresponding  to  this  is  (500  —  144)  41  = 
14,596  Ibs.  per  sq.  in. 

F  =  2  tJtR8Ci  =  2  X  0.1225  X  15,000  X  48  X  1.25 

=  220,000  Ibs. 
M  =  F  X  1.56  X  R8  =  220,000  X  1.56  X  48  =  16,500,000  in.  Ibs. 

The  inertia  of  the  section  is 

/  =  A  (D2  +  d2)  +  0.394  X  n  X  £A 
.  10 

=  H??5  (io22  +  882)  +  0.394  X  15  X  963  X  0.1225, 
I  =  2,365,000  +  640,000  =  3,005,000, 

M  JI_  _  72  X  3,°o5,°oo  . 
e  51 


CHIMNEYS  269 

The  remaining  moment  to  be  resisted  by  the  chimney  is 

19,570,000  -  4,240,000  =  15,330,600  in.  Ibs. 
F  =  M  -4-  (1.56  X  R8)  =  15,330,600  -T-  (1.56  X  48)  =  204,000  Ibs. 
ft  =  F+(2XtaX  £.'XC,)  =  204,000  -r-  (2  X  0.1225  X  48  X  1.25) 

=  13,800  in.  Ibs. 

,        11,800 

fc  =  ^~   -  =  340  Ibs. 
41 

The  extreme  fiber  stress  will  be  somewhat  in  excess  of  this. 

k  =  0.46,    A  =  k  X  Rs  =  0.46  X  48  =  22  ins. 

By  similar  triangles  the  extreme  fiber  stress  is  -         ^  X  340 

=  378  Ibs.  per  sq.  in.  The  extreme  fiber  stress  then  becomes 
(2  X  72)  +  378  =  522  Ibs.  per  sq.  in.  The  reinforcing  bars 
will  require  an  area  of  2085  X  0.0175  =  36-5  scl-  ms-  Using 


f-in.  0  bars  will  demand   -     =  76  —  f-in.  0  bars. 

0.44 

Similar  calculations  will  indicate  the  following  bars  in  the 
several  sections;  beginning  at  the  bottom,  we  have 

Section.  Section. 

First,  22  ft.,  76  f-in.  round  bars.  Fourth,    22  ft.,  28  f-in.  round  bars. 

Second,  25  ft.,  60  f-in.  round  bars.  Fifth,       34  ft.,  14  f-in.  round  bars. 

Third,  22  ft.,  44  f-in.  round  bars. 

Horizontal  reinforcing  rings  must  be  used  to  resist  the  web  stresses 
and  also  to  prevent  cracking  due  to  expansion  from  the  heat. 
Concrete  being  a  poor  conductor  of  heat  the  inside  becomes  much 
hotter  than  the  outside.  This  causes  considerable  circum- 
ferential stress  in  the  chimney  which  must  be  resisted  by  hori- 
zontal rings,  preferably  placed  near  the  outer  surface  of  the 
chimney. 

The  amount,  of  this  circumferential  reinforcement  can  be  only 
roughly  approximated,  as  the  difference  in  temperature  between 
the  inner  and  outer  faces  of  the  wall  must  be  guessed.  In 
several  chimneys  this  reinforcement  was  from  J  to  f  of  i  per  cent. 


270  GRAPHICS  AND   STRUCTURAL  DESIGN 

The  greater  the  steel  ratio  the  higher  is  the  compression  in  the 
concrete. 

These  chimneys  are  generally  lined  at  least  one-third  their 
height.  The  inner  tubes,  being  protected  from  the  wind,  are 
merely  called  upon  to  resist  temperature  stresses.  These  are 
not  usually  calculated.  The  linings  may  be  made  of  fire  brick 
or  reinforced  concrete  and  will  have  horizontal  ring  reinforce- 
ment somewhat  lighter  than  in  the  outer  wall,  say  from  £  to  J 
of  i  per  cent,  and  vertical  reinforcement  to  resist  temperature 
stresses  of  from  J  to  ^  of  i  per  cent. 

The  horizontal  reinforcing  rings  are  commonly  ^-in.  or  f-in. 
rounds,  and  are  spaced  from  12  ins.  to  18  ins.  vertically.  The 
spacing  should  be  closer  at  the  point  on  the  chimney  where  the 
lining  stops,  Fig.  286,  and  additional  vertical  reinforcement  should 
also  be  placed  at  this  part. 

Chimney  Base,  Fig.  288.  —  Estimate  of  total  weight  of 
chimney : 

Section  Area,  Volume,  Weight, 

sq.  ft.  cu.  ft.  Ibs. 

Upper  iffi-  X  85  =  715  X  144  =  103,000 

Lower  -2T°?Y-  X  40  =  580  X  144  =    83,500 

Lining  ff \  X  40  =  265  X  144  =    38,200 

Base  56  X    3  =  168  X  144  =    24,190 

Foundation  (341  +  486)  X  144  =  167,000 


Total        415,890 

The  total  wind  moment  previously  found  for  the  base  of  the 
chimney  is  1,634,550  ft.  Ibs.  The  total  force  is  (85  X  6.8  X  30) 
+  (40  X  8.5  X  30)  =  27,540  Ibs.  The  resultant  wind  pressure 

then  must  act-1'  34>55°  =  50.5  ft.  above  the  ground  line.     The 
27,540 

resultant  of  weight  and  wind  will  act  a  distance  q  from  the  axis 
of  the  chimney  and 


CHIMNEYS  271 

The  radius  of  the  kern  for  a  square  is  r  =  0.118  X  h  =  0.118 
X  20  X  12  =  28.4  ins.     From  page  246, 


/2  2O 

since 

TJ/  r      r>2  W 

r  =  «./..^,  or  /e  =  _  =  _ 

0  =  0.25.  The  resultant  pressure  passes  4.4  ft.  from  the  cen- 
ter so  that  the  pressure  extends  a  distance  (5.6  X  3)  =  16.8  ft. 
over  the  base.  The  area  of  the  pressure  on  the  base  is  16.8  X  20 
=  336  sq.  ft.  The  total  pressure  equals  the  entire  weight  of 

the  chimney,  hence  336  X*~  =  415,890    and 

p  =  (415,890  X  2)  -5-  336  =  2470. 

The  reinforcement  can  be  only  approximated,  the  simplest  way 
being  to  consider  the  part  cdef,  Fig.  288,  as  a  can  tile  vered  beam. 
The  distance  from  the  center  of  gravity  of  the  portion  cdef  to 
the  line  cf  is 

Area.  Statical  moment. 

+  20  X  V-  =  ioo  X  -V-  =333-33 

-ioX    |=    f|X  (5  +  I)  =166.75 

166.58 


75 

The  load  on  cdef  approximates  75  X  2400  =  180,000  Ibs.  The 
approximate  bending  moment  is  M  =  180,000  X  (5  —  2.22) 
=  500,400  ft.  Ibs.  The  bending  moment  per  foot  along  the 
line  df  is  (500,400  X  12)  -5-  10  =  600,480  in.  Ibs.  Assuming 
n  =  15;  d  =  42  ins.  and  p  '=  0.0025;  j  =  0.922  and  kj  =  0.218. 
Using  the  formula  for  reinforced  concrete  design,  we  have 

-  Ma  _    _  600,480  _ 

**      A  Xjd  ~  (42  X  12  X  0.0025)  X  0.922  X  42 
=  1  2,  300  Ibs.  per  sq.  in. 


272  GRAPHICS  AND   STRUCTURAL  DESIGN 

and 

,        2  XMC  2  X  600,480 

fc  =  r^ — r~^  =  —  ;  =  260  Ibs.  per  sq.  in. 

kj  X  bd2      0.218  X  12  X  422 

The  depth  of  the  beam  being  controlled  by  the  necessity  of  the 
foundation  reaching  proper  soil  and  extending  below  the  frost 
line  this  fiber  stress  in  the  concrete  is  satisfactory.  The  spacing 
of  i|-in.  round  bars  will  be 

area  of  bar  X  12  0.994  X  12  ,  . 

spacing  = =  — — -^  — =  9f  ins. 

A  42X12  X  0.0025 


CHAPTER   XVII 
RETAINING  WALLS 

PRESSURES  ON  RETAINING  WALLS 

ACCORDING  to  Coulomb's  theory  some  wedge  BA C,  in  Fig.  289, 
will  produce  a  maximum  pressure  E  against  the  wall.  This 
force  will  make  an  angle  5  with  the  face  of  the  wall  corresponding 


1 


FIG.  289. 


to  the  angle  of  friction  between  the  face  of  the  wall  and  the  fill. 
The  natural  slope  of  the  fill  is  the  angle  <£;   the  angle  that  the 
face  of  the  wedge,  producing  the  maximum  pressure  against  the 
wall,  makes  with  the  horizontal  is  x.' 
h  =  height  of  wall  in  feet. 

w  =  the  weight  of  the  fill  in  pounds  per  cubic  foot. 
According  to  this  theory  the  general  value  of  the  maximum 
pressure  in  pounds  per  foot  of  length  of  wall  is 


sin'  •  •  sin  (•  + 


(x  +  V/S!" 
\        V  sm 


+  ;>  5!n  <*  ~  " 
+  S)sm  (0  -a 


273 


274 


GRAPHICS  AND   STRUCTURAL  DESIGN 


This  is  simplified  for  the  more  commonly  assumed  conditions 
as  follows: 

For  6  =  90  degrees,  and  5  =  a, 


E  =  -wh2 

2 


COS 


sin  (0  +  a)  sin  (0  —  a 


In  Rankine's  formula  0  =  90°,  a  =  o,  and  5  =  o;  substituting 
these  values  in  the  general  equation  gives: 

„      i     72      cos20 

E  =  ~wh2-  —  -  —  r^-r2- 

2  (l  +  SH10)2 

E  =  \w  .  A2  tan2  (45°  -  10). 


2  i  +  sin  0 

This  theory,  although  usually  developed  for  retaining  walls, 
applies  also  to  the  sides  of  bins. 


FIG.  290. 

Graphical  Solution.  —  The  graphical  method  (Fig.  290)  affords 
the  neatest  way  of  estimating  the  pressure  E. 

E  =  %w*p.y. 


RETAINING   WALLS  275 

p  and  y  =  lengths  of  lines  in  the  diagram  measured  to  the  same 
scale  by  which  the  wall  or  side  is  drawn.  The  other  symbols 
are  the  same  as  in  the  preceding  case. 

Explanation  of  Diagram.  —  A  B  is  the  side  of  the  wall  or  bin 
in  contact  with  the  fill.  A  I  is  a  horizontal  line  through  the  base 
of  the  wall  or  side  AB.  AC  is  the  line  of  natural  slope  of  the 
fill  making  an  angle  <f>  with  the  horizontal.  BC  is  the  slope  of 
the  top  of  the  fill  making  an  angle  a  with  the  horizontal.  Pro- 
duce AC  and  BC  until  they  intersect  in  C.  Upon  AC  as  a  diam- 
eter draw  the  semicircle  ARC.  Draw  BG  making  an  angle 
0  +  6)  with  .the  wall  AB. 

At  the  point  of  intersection  G  of  BG  and  AC  draw  GH  perpen- 
dicular to  AC.  With  AH  as  a  radius  draw  the  arc  HF  and 
through  F  draw  DF  parallel  to  BG.  From  D  drop  p  perpendicu- 
lar to  AC. 

The  proofs  and  discussions  of  the  preceding  equations  and 
diagrams  may  be  found  in  books  on  the  mechanics  of  retaining 
walls  and  earth  pressures. 

Figure  291  is  the  diagram  when  the  angle  of  fill  a  equals  the 
angle  of  natural  slope  <J>.  In  this  case  since  BK  and  AF  are 
parallel  p  and  y  will  be  the  same  length  wherever  the  point  D 
is  taken  in  the  line  BK. 

Figure  292  shows  the  construction  when  the  angle  of  fill  falls 
below  the  horizontal  and  also  below  the  line  BC1  making  an  angle 
</>  -f  6  with  the  back  of  the  wall.  In  all  the  preceding  cases 
E  =  |  w  •  p  -y. 

If  the  fill,  Fig.  293,  carries  a  uniform  load  L  Ibs.  per  sq.  ft.  of 
horizontal  projection,  the  height  of  the  fill  may  be  considered 
as  increased  sufficiently  to  bring  such  loading  on  the  soil.  The 
diagram  is  given  in  Fig.  294. 


w'  =  w  +  ?——     and    E  =  -\w  + 


Distribution  of  Pressure  on  the  Wall  or  Side.  —  In  the  first 
cases,  where  the  fill  is  not  loaded,  the  pressure  on  the  wall  will 


276 


GRAPHICS  AND  STRUCTURAL  DESIGN 


r 


i: 


FIG.  293. 


FIG.  294. 


FIG.  296. 


FIG.  297. 


RETAINING   WALLS 


277 


vary  as  a  triangle  and  the  center  of  pressure  will  correspond  to 
the  center  of  gravity  of  the  triangle,  as  shown  in  Fig.  295. 

Where  the  fill  is  loaded  the  usual  assumptions  regarding  the 
distribution  of  the  pressures  behind  the  wall  are  shown  in  Fig.  296 
and  the  magnitudes  of  these  pressures  are  given  by  the  follow- 
ing formulae.  The  resultant  pressure  Et  acts  through  the  center 
of  gravity  of  the  trapezoid  of  pressure  abed. 

2-L\     .  ,     „       /E'  +  E"\h 


Et  =  l 


and 


From  which 


hi 


Center  of  Gravity  of  a  Trapezoid.  —  Fig.  297  being  a  trape 
zoid  its  center  of  gravity  can  be  readily  found  as  shown. 

WEIGHTS  AND  ANGLES  OF  REPOSE  OF  MATERIALS 


Material. 

Weight  per  cu.  ft. 

Angle  of  repose,  <t>. 
Degrees. 

Clav.  dry 

QO—IIO 

^0—4.0 

Clay,  damp 

IOO—  I2O 

40—4? 

Clay,  wet 

I  2O—  I7,<» 

1C—  2C 

Gravel,  wet  

IOO—  I2O 

2^-40 

Ashes  

40—4.5 

25—40 

Coke  

•?o 

30-4? 

Earth,  dry 

80-90 

20—40 

Earth,  moist 

9O—IOO 

7.5—45 

Earth,  wet 

IO5—I2O 

17—30 

Broken  stone,  wet.  .  . 

IOO 

7  {{—40 

Coal,  broken  

56 

45  —  5O 

Sand,  dry  

90-110 

7Q—  35 

Sand,  moist  

IOO-IIO 

70—45 

Sand,  wet 

IIC—  I2S 

15—  7Q 

Water 

62  q 

o 

The  coefficient  of  friction  between  the  usual  masonry  materials 
upon  themselves  or  upon  soil  will  range  from  0.50  to  0.75. 

RETAINING  WALL 

A  retaining  wall  may  fail  by  being  rotated  about  the  toe  A 
or  by  sliding  upon  the  base  AB.  There  is  usually  little  likeli- 
hood of  failure  by  sliding  as  the  coefficient  of  friction  between  the 


278 


GRAPHICS  AND   STRUCTURAL   DESIGN 


wall  and  the  soil  is  high.  The  force  F  due  to  the  earth  pressure 
behind  the  wall  creates  an  uneven  pressure  under  AB  as  shown 
in  Fig.  298. 

A  tilting  of  the  wall  may  result  from  the  side  at  A  settling 
faster  than  that  at  B.  To  improve  the  condition  it  is  not 
necessary  to  enlarge  the  entire  wall  but  the  base  may  be  spread 
as  shown  in  Fig.  299,  so  that  the  resultant  R  passes  through  the 
center  of  the  base  AB. 


D        C 


FIG.  298. 


FIG.  299. 


At  best,  the  theories  relating  to  retaining  walls  are  unsatis- 
factory. This  is  due  largely  to  the  fact  that  the  properties  of  the 
materials  vary  so  widely  that  it  is  difficult  to  decide  upon  the  con- 
ditions applicable  to  a  particular  case.  The  practical  conditions 
differ  greatly  from  the  approximations  necessary  in  the  theory. 
The  weight  of  the  fill  and  its  angle  of  repose  will  vary  greatly 
as  the  fill  is  wet  or  dry,  clean  or  dirty,  loose  or  rammed,  etc. 

Other  factors,  such  as  shock,  unexpected  loading  and  frost, 
cannot  be  taken  into  account  in  the  formulae.  According  to 
Trautwine  a  practically  vertical  retaining  wall  sustaining  a  fill 
of  sand,  gravel  or  earth,  without  surcharge,  with  the  fill  loose, 
not  rammed,  the  wall  being  of  good  common  rubble  or  brick, 
should  have  a  thickness  at  the  top  of  the  footing  of  four-tenths 
the  height  of  the  wall.  The  Engineering  News  in  its  issue  of 


RETAINING  WALLS  279 

Sept.  26,  1912,  commenting  upon  the  failure  of  a  retaining  wall, 
recommends  the  calculation  of  a  retaining  wall  upon  the  assump- 
tion that  a  wall  with  no  surcharge  carries  a  load  at  least  equiva- 
lent to  a  mass  of  water  behind  the  wall  of  two-thirds  its  depth. 
The  following  problem  will  illustrate  the  method  of  designing  a 
retaining  wall. 

Problem.  —  Design  a  section  for  a  retaining  wall,  height  20  ft., 
weight  of  earth  no  Ibs.  per  cu.  ft.,  natural  slope  30  degrees, 
maximum  earth  pressure  not  to  exceed  5000  Ibs.  per  sq.  ft. 
Assume  that  the  concrete  of  the  wall  weighs  140  Ibs.  per  cu.  ft. 

The  center  of  gravity  of  the  wall  must  be  located  with  reference 
to  the  back  of  the  wall.  Dividing  the  wall  section  (see  Fig. 
300)  into  rectangles  and  triangles,  multiplying  their  several 
areas  by  the  distances  of  their  centers  of  gravity  from  the  back 
of  the  wall  and  then  dividing  the  sum  of  these  moments  by  the 
total  area  of  the  wall  section  gives  the  center  of  gravity  of  the 
entire  wall  section  as  46.6  ins.  from  the  back  of  the  wall.  The 
area  of  the  wall  section  is  143  sq.  ft. 

The  weight  of  i  ft.  of  length  of  the  wall  is  143  X  140  =  20,020 
Ibs.  The  resulting  earth  pressure  is 

E  =  %-w-p-y  =  |  X  no  X  12.2  X  14  =  9394  Ibs. 

The  resultant  R  passes  9.3  ins.  to  the  left  of  the  center  line  a-a 
of  the  base.  Using  the  formula  deduced  for  footings  with  vary- 

W 

ing  pressures  on  the  soil,  page  232,  fe  =  j-jr2  (B  —  6#),  here 

W  =  25,000  Ibs.  and  is  the  vertical  component  of  E  and  the 
weight  of  the  wall,  b  =  i  foot,  and  the  maximum  pressure  under 
the  foundation  is 

6 


Since 


An  inspection  of  the  wall  will  show  that  by  increasing  the 
width  of  the  footing  to  12  ft.,  as  shown  in  Fig.  301,  the  resultant 


280 


GRAPHICS  AND  STRUCTURAL  DESIGN 


would  be  brought  practically  through  the  center  of  the  base  thus 
making  the  pressure  uniform  across  it. 

To  prevent  sliding  the  coefficient  of  friction  would  have  to  be 
8100  -5-  25,000  =  0.324.  As  the  coefficient  of  friction  probably 
lies  between  0.500  and  0.750  there  is  evidently  ample  margin  of 
safety  against  failure  in  this  way. 


8100* 


FIG.  300. 


REINFORCED-CONCRETE  RETAINING  WALLS 

The  use  of  reinforced  concrete  in  retaining  walls  has  led  to 
some  modification  of  the  ordinary  design.  This  results  in  con- 
siderable economy  in  walls  exceeding  25  ft.  in  height.  The  com- 
mon section  of  such  a  wall  is  shown  in  Fig.  302.  The  vertical 
load  upon  the  soil  is  the  weight  of  the  wall  plus  the  weight  of  the 
soil  carried  directly  by  the  wall,  that  is,  the  soil  prism  abed  and 
the  vertical  component  of  the  force  E  acting  upon  the  back  of 
the  wall.  The  general  lines  of  the  reinforcement  are  indicated 
by  the  steel  shown  in  Fig.  302.  The  buttresses  B  are  placed  at 
intervals  of  from  8  to  10  ft.  along  the  wall.  Walls  under  18  ft. 


RETAINING   WALLS 


281 


k — |000B- >| 


282 


GRAPHICS  AND   STRUCTURAL  DESIGN 


may  have  the  buttresses  omitted  and  the  wall  designed  as  a 
cantilever  beam. 

Problem.  —  Design  a  reinforced-concrete  wall  for  a  height  of 

25ft. 

Assume  the  weight  of  the  fill  as  100  Ibs.  per  cu.  ft.  The  angle 
of  repose  of  the  material  behind  the  wall  can  be  taken  as  30  de- 
grees. The  maximum  soil  pressure  is  not  to  exceed  2  tons  per 
sq.  ft.  The  working  unit  stress  in  the  steel  is  not  to  exceed 
12,000  Ibs.  per  sq.  in.,  while  that  in  the  concrete  is  not  to  be  over 
500  Ibs.  per  sq.  in. 

The  fill  abed,  Fig.  303,  in  the  wall  will  act  to  prevent  over- 
turning, so  that  we  will  first  want  the  weight  and  the  center  of 
gravity  of  the  wall,  buttresses  and  prism  abed  of  the  fill. 


Section. 

Volume.               Weight. 

Arm. 

Moment. 

Coping 

25X140=     3>5oo 

9-  33 

32,670 

Face  of  wall  .    . 

217.5X140=  30,450 

9.50 

289,275 

Fillet     .      .           

7.5X140=     1,050 

IO.  25 

10,760 

Fillet  .  

7.5X140=      1,050 

10.83 

11,372 

Base  

280X140=   39,200 

7.00 

274,400 

Base  

15X140=     2,100 

0.75 

1,575 

Buttress  

128.1X140=   17,934 

5.55 

99,534 

Buttress 

22    9X140=       3>2O6 

8.67 

27,785 

Fill,  prism  over  base 

I7C9     CX  IOO=I7^,  9^O 

4.50 

791,775 

Fill,  prism  over  buttress  

160.0X100=   16,000 

3.00 

48,000 

Total  for  10  ft.  length 

290,440 

1,587,146 

Distance  to  center  of  gravity  of  wall,  x  =    ^  ''        =  5.46  ft. 

290,440 

E  =  I  (w  •  p  .  y)  =  \  (100  X  14-43  X  1443)  =  10,400  Ibs. 

For  a  length  of  10  ft.  this  is  104,000  Ibs. 

The  diagram  made  to  an  enlarged  scale  of  Fig.  303  gives  the 
line  of  action  of  the  resultant  of  E  and  the  total  weight,  290,440 
Ibs.,  as  passing  through  a  point  35.8  ins.  to  the  left  of  the  vertical 
line  through  the  center  of  gravity.  The  distribution  of  this 


RETAINING   WALLS  283 

pressure  upon  the  foundation  is  given  by  the  formula 


=  14,    *  =          =  j  44  ft 

12 


and  falls  to  the  right  of  the  center,  hence 


Sliding  of  the  Wall.  —  The  coefficient  of  friction  between  the 
wall  and  the  soil  will  lie  between  50  per  cent  and  75  per  cent. 
The  resistance  to  sliding,  even  assuming  the  coefficient  at  50  per 
cent,  would  be  290,440  X  0.50  =  145,220  Ibs.  This,  being  con- 
siderably above  the  horizontal  pressure  on  the  wall,  104,000  Ibs., 
indicates  little  possibility  of  failure  by  sliding. 

REINFORCEMENTS 

Face  of  the  Wall.  —  The  wall  will  be  tied  into  the  buttresses 

W  *L 

and  the  bending  will  be  assumed  as  M  =  -     —  •     The  reinforce- 

12 

ments  will  be  determined  for  sections  down  the  wall  and  the 
bending  moment  will  be  calculated  from  the  load  taken  from 
Fig.  304.  As  the  pressure  is  assumed  as  varying  proportionally 
to  the  depth,  the  distribution  will  be  represented  by  a  triangle. 
The  area  of  this  triangle  equals  104,000  Ibs.,  or 

104,000  =  0.5  X  ac  X  ab  or  ab  =  (104,000  X  2)  -s-  25  =  8320  Ibs. 

The  load  upon  a  strip  i  ft.  wide  and  10  ft.  span  is  then  given 
by  an  intercept  in  the  triangle  abc  parallel  to  ab,  the  intercept 
being  located  the  same  distance  from  c  that  the  center  line  of  the 
section  is  from  the  top  of  the  wall;  thus  the  load  on  a  strip 
12  ins.  wide  whose  center  is  10  ft.  from  the  top  is  3450  Ibs. 
when  scaled  from  Fig.  304. 


284  GRAPHICS  AND  STRUCTURAL  DESIGN 

W  •  L 
The  bending  moment  upon  such  a  strip  is  M  =  -      -  =  3450 

X  10  X  xf  =  34, 500  in.  Ibs.     Using  the  approximate  formulae 

Ms  =  J  X^  Xf8  Xd. 
Taking          d  =  10.5  ins.  and/8  =  14,000 

34,500  X  8 

makes  A  = i2-LLi2 =  0.296  sq.  in. 

7  X  14,000  X  10.5 

f-in.  0  bars  would  require  a  spacing  of 

Area  of  i-in.  0  X  12      0.196  X  12  .  _  . 

=  — - — - — .  approximately  8  ins. 

0.296 

If  desired,  this  can  be  checked  by  the  more  accurate  formula. 

0.296 

p  = —  =  0.0024. 

10.5  X  12 

p  *n  —  0.00235  X  15  =  0.035.      From    the   curves  j  =  0.924; 
k-j  =  0.231. 

Substituting  these  values  in  formulae  (3)  and  (4) 

Ma  =  A  Xf3  Xj'd    or    /. 


0.296  X  0.924  X  10.5 


AXJ-d 

=  12,000  Ibs. 


r  2  Mc  2    X   34,500 

and      fc  =  7  -^  =  -  T  =  225  Ibs. 

k-jXb.d*      0.231  X  12  X  io.52 

These  values  are  satisfactory,  it  not  being  desired  to  make  the 
wall  thinner.  The  J-in.  round  bars  will  be  spaced  8  ins.  center 
to  center  in  the  section  between  8  ft.  and  10  ft.  from  the  top  of 
the  wall. 

The  reinforcements  may  be  determined  for  the  other  sections 
in  a  similar  way.  The  results  of  such  calculations  are  given  in 
the  table.  The  calculated  spacing  is  given  in  the  brackets,  while 
that  to  be  used  is  given  outside. 


RETAINING   WALLS 


285 


Depth  of 
section, 
feet. 

Load, 
pounds. 

-•^- 

inch  pounds. 

Area  of  steel, 
square  inches. 

Size  of  bars  and  spacing, 
inches. 

4 

1400 

14,000 

o.  109 

|  "  0  spaced  (21.6)  12" 

7 

2400 

24,000 

0.187 

¥  0  spaced  (12.6)  12" 

IO 

345° 

34,500 

0.296 

£"0  spaced  (8     )    8" 

12 

4100 

41,000 

0.318 

-2-"0  spaced  (7.4)    7-5" 

14 

4800 

48,000 

0-373 

\"  0  spaced  (  6.3)    6" 

16 

5400 

54,000 

0.420 

\"  0  spaced  (5.6)    5.5" 

18 

6100 

61,000 

0.474 

\"  0  spaced  (4.96)    5.0* 

20 

6750 

67,500 

0.524 

\"  0  spaced  (  4.5)    4.5" 

22 

7400 

74,000 

0-575 

\"  0  spaced  (  4.1)    4.0" 

The  toe  or  base  of  wall  to  the  left  of  the  face  will  carry  a  load 


given  in  Fig.  305  and  equals 


=  2975  Ibs.  per  sq.  ft., 


making  a  load  per  foot  of  wall  of  2975  X  4  =  11,900  Ibs.  The 
distance  of  the  center  of  gravity  of  this  loading  from  the  face  of 
the  wall  is 

x  =  3350  +  2  (2600)  x  48  =       lbs 

3350  +  2600         3 

The  moment  on  a  strip  12  ins.  wide  then  is  M  =  11,900  X  25  = 
297,500  in.  lbs. 

By  the  approximate  formula  Ma  =  |  X  A  X  /.  X  d. 
A  =      297,500  X  8 
7  X  14,000  X  22 


1.07  sq.  ins. 


The  spacing  for  i-in.  0  bars  is 


0.785  X  12 
1.07 


=  8.75  ins. 


Rear  Portion  of  Base.  —  The  load  upon  this  section  (see  Figs. 
305  and  306)  will  be  the  difference  between  the  pressure  upon  the 
soil  and  the  weight  of  the  nil  vertically  over  this  portion.  This 
amounts  to  23  ft.  of  soil  and  2  ft.  of  concrete,  and  weighs 
2300  +  280  =  2600,  approximately.  The  maximum  loading  is 
2600  —  800  =  1800  Ibs.  Estimating  the  reinforcement  for  maxi- 
mum loading  the  total  load  on  a  strip  12  ins.  wide  and  10  ft. 
long  is  18,000  Ibs. 

18,000  X  io  X  12 
12 


M 


=  180,000  in.  Ibs. 


286  GRAPHICS  AND   STRUCTURAL   DESIGN 

Placing  rods  3  ins.  above  the  bottom  of  the  slab  and  using  the 
approximate  formula  gives, 

A        8  XM  8  X  180.000 

A  =  -         —  = -  =  0.70  sq.  in. 

jXp-d      7X14,000X21 

rnu  •  r    7   •         r*   u  MI   u     °-6o  X  12 

The  spacing  of  --in.  0  bars  will  be  -  -  =  ioms.,  approx- 

8  0.70 

imately. 

W  •  L  W  •  L 

In  assuming  the  bending  moment  at  If  =  -     —  instead  of  — — 

12  8 

it  becomes  necessary  to  place  reinforcement  in  the  outer  flange 
over  the  supports;  this  would  have  to  equal  at  least  50  per  cent 
of  the  reinforcement  at  the  middle.  Here  short  lengths  of  rods 
the  same  as  the  full-length  reinforcements  (see  Fig.  306)  will  be 
placed  over  the  supports  and  spaced  the  same  as  the  other  rods. 
Counterforts.  —  In  this  design  (see  Fig.  302)  these  will  resist 
the  moment  due  to  the  thrust  on  the  wall.  The  thrust  on  the 
wall  from  the  top  to  the  upper  face  of  the  wall  is  -%5—  X  23 
=  87,975  Ibs. 

23   X  12 

The  moment  is   88,000  X  -          -  =  8,096,000  in.  Ibs. 

o 
The  design  can  be  made  as  a  T  beam,  the  wall  serving  as  the 

flange. 

The  distance  from  the  face  of  the  wall  to  the  center  of  the 
reinforcement  can  be  assumed  as  108  ins.  The  ratio  of  flange 
thickness  to  depth  of  beam  is  -j1^-  =  o.n.  From  the  curves  for 
T  beams,  j  =  0.95,  and  substituting  in  the  equation 

M  =  A  X/8  Xj*d, 
we  have 

M  8,096,000 

A  = : — :  =  -^-^ =  5.65  sq.  ins. 

faXj-d      14,000  X  0.95  X  108 

This  will  require  eight  yf-in.  0  rods. 

The  number  of  rods  can  be  reduced  towards  the  top  of  the 
wall  as  the  bending  moment  becomes  less. 


RETAINING   WALLS 


287 


In  a  beam  similar  to  the  counterfort  the  flange  force  at  any 
distance  x  from  the  top  can  be  approximately  expressed  as 

(F  •  x2} 
F  •  x  =  v  ,  where  h  =  height  of  wall  whose  flange  force  is  F. 

This  is  not  accurate,  as  j  will  vary  with  the  depth  of  the  beam, 
but  may  be  approximated  at  T90.  The  flange  forces  will  be  pro- 
portional to  the  required  areas.  The  varying  flange  forces  can 
then  be  represented  by  laying  off  a  base  line  equal  to  the  height 
of  the  wall  and  plotting  ordinates  according  to  the  formula  just 
stated.  The  curve  being  a  parabola  can  be  laid  off  as  in  Fig.  307. 

ca  represents  the  side  of  the  wall,  here  25  ft.,  and  cd  represents 
the  side  extending  to  the  top  of  the  slab,  de  is  the  force  in  the 
steel  23  ft.  from  the  top  of  the  wall,  gc  represents  the  total 
length  of  diagonal  steel  reinforcements,  the  distance  df  measuring 
8  ft.  from  the  wall. 

Eight  rods  taken  in  pairs  will  divide  de  into  four  equal  parts 
and  dh,  hi,  ij  and  jb  each  represents  the  area  of  two  of  the  rods. 
To  find  the  lengths  of  the  shortest  rods 
produce  jk  perpendicularly  until  it 
cuts  the  curve,  from  k  project  a 
horizontal  line  until  it  cuts  the  diag- 
onal line  in  /.  The  lengths  of  these 
two  rods  will  then  be  given  by  the 
length  gl.  As  this  curve  gives  the 
rate  of  change  of  flange  force  it  also 
gives  the  horizontal  shear  which  will 
be  the  difference  between  two  adja- 
cent abscissas  and  the  stirrups  can  be 
determined  as  shown  for  beams  with 
uniform  loads,  page  211.  The  beam 
being  so  deep  the  concrete  alone 
would  probably  supply  ample  strength 
for  horizontal  shear  but  in  practice 
there  are  generally  also  metal  stirrups  inserted. 

Another  type  of  wall  is  the  can  tile  vered  wall.     In  it  the 


288  GRAPHICS  AND  STRUCTURAL  DESIGN 

buttresses  or  counterforts  are  omitted.  This  type  is  suited  to 
lower  walls  than  the  design  just  carried  out.  The  principal 
reinforcements  are  shown  in  Fig.  308:  The  calculation  of  the 
thicknesses  and  reinforcing  metal  can  be  made  on  the  same 
general  lines  as  the  preceding  problem. 

The  vertical  wall  CD  is  treated  as  a  cantilever  and  is  assumed 
as  carried  by  the  base  AB. 


CHAPTER   XVIII 
BINS 

PRESSURES  ON  BINS 

THE  'methods  used  in  obtaining  the  pressures  upon  retaining 
walls  are  also  applicable  to  the  pressures  acting  on  the  sides  and 
bottoms  of  bins. 

Bins  may  serve  as  retainers  of  any  material  but  the  discussions 
here  will  be  confined  to  such  bins  as  are  ordinarily  used  for  coal, 
sand,  refractory  materials  at  steel  mills,  and  similar  materials. 
Bins  at  grain  elevators  are  generally  high  compared  with  their 
width;  it  therefore  follows  that  the  friction  of  the  material  against 
the  sides  of  such  bins  may  greatly  assist  in  relieving  the  lower 
portion  of  the  bin,  both  sides  and  bottom,  of  this  accumulative 
pressure.  The  theory  previously  given  for  retaining  walls  does 
not  apply  in  this  case  as  the  assumed  plane  of  rupture  to  produce 
maximum  pressure  will  not  cut  the  surface  of  the  grain.  Several 
formulae  have  been  proposed  for  estimating  these  pressures  ;  that 
of  Janssen,  taken  from  Hiitte  des  Ingenieurs  Taschenbuch,  is 


and  L  =  V-k. 

V  =  vertical  pressure  of  the  grain,  pounds  per  square  foot. 

/  =  coefficient  of  friction  of  grain  on  the  bin  wall. 

w  =  weight  of  grain,  pounds  per  cubic  foot. 

R  =  hydraulic  radius  of  horizontal  section  of  bin.     For  circu- 

,.       „      diameter      ,.,          ,  .         D 

lar  section  R  =  --     For  other  sections  R  = 


4 
area  of  section 


perimeter  of  section 

289 


All  dimensions  in  feet. 


290  GRAPHICS  AND   STRUCTURAL  DESIGN 

k  =  ratio  of  lateral  to  vertical  pressure. 
e  =  base  of  Naperian  system  of  logarithms. 
h  =  depth  to  bottom  or  point  on  side  at  which  pressure  is  to 
be  determined. 

For  wheat,  w  =  50  Ibs.  per  cu.  ft.;  k  =  0.60;  /  =  0.40. 


SHALLOW  BINS 

The  bins  now  taken  up  will  be  shallow  compared  with  those 
just  considered.  In  these  the  theory  of  retaining  walls  will  apply. 
These  bins  will  commonly  be  made  of,  (a)  Timber,  (b)  Steel, 
unlined,  or  lined  with  timber  or  concrete,  (c)  Reinforced  concrete. 

The  bin  should  be  built  or  lining  placed  to  avoid  pockets  that 
might  hold  materials  indefinitely.  This  is  particularly  neces- 
sary in  materials  in  which  spontaneous  combustion  might  occur. 
Timber  lining  should  be  tar  coated  on  the  sides  and  faces  against 
the  metal,  when  such  treatment  will  not  affect  the  material  held. 

The  inclined  sides  of  unlined  steel  bins  frequently  have  wear- 
ing strips,  flats  about  3  X  Jin.,  placed  on  about  i2-in.  centers 
and  running  at  right  angles  to  the  direction  of  flow  of  the  material, 
thus  tending  to  retain  a  slight  thickness  of  the  contained  material 
along  the  side  or  make  the  material  slide  upon  itself  rather  than 
on  the  metal  and  so  protect  the  metal  from  wear. 

STRESSES  IN  BESTS 

Although  the  forces  acting  upon  the  bin  sides  may  be  esti- 
mated by  formulae,  the  graphical  methods  used  on  retaining 
walls  will  be  largely  used  in  the  following  discussions.  For  those 
who  prefer  formulae  those  given  by  R.  W.  Dull  in  the  Engineering 
News  of  July  21,  1904,  are  here  given,  with  slight  modification. 


BINS 


291 


where 


\n  +  i  /    2  cos  5 


__      /  cos  0  \2  wh2 

IV   =  —  > 

\«  +  I  /        2 


+  5)  sin 


cos  5 


J:_MI_.J A 


FIG.  309.     Vertical  wall,  no  surcharge. 


where 
If 


_w^_      /cos^y 
2  cos  5       \i+n     ' 


!    2 


_  *  /sin  (</>  -f-  5)  sin  (<#>  —  a) 
V  cos  6  cos  a 

a  =  <f>,     then     w  =  o. 


FIG.  310.     Vertical  wall,  surcharged. 


292 


GRAPHICS  AND  STRUCTURAL  DESIGN 
E  = 


where 


_  4  /sin  (<f)  +  5)  sin  (</>  +  a) 
V  cos  5  cos  a 


N-* 
E 


J. 


FIG.  311.    Vertical  wall,  surcharge  negative. 


where 


wh* 


2cos08  +  5) 


X 


= 


cos  Q3  -  ^>)  "I2 

' 


2  cos  03  +  5)         (w  +  i)cos/Sj 


+  5)  sin  </> 


cos  (5  +/S)cos/3 


FIG.  312.    Wall  inclined  outward.     6  <  90°  +  5.     Surface  horizontal. 


BINS 


293 


where 


E  = 


wh* 


2058(5+0) 


X 


i)cos/3 


_     wh2  cos  5         ["cosQfr  -/3)  "I8 
2cos(S+/3)  X  L(w  +  i)  cos/3  J  ' 


sin  (</>  +  5)  sin  (<j>  —  a) 
cos  (5  +  0)  cos  (ft —  a) 


FIG.  313.     Wall  inclined  outward.    6  <  90°  -f  5.     Surcharged. 


IF  =  weight  of  triangle  ABC  = 
E  =  — 


(45<>-f) 


tan 


i  E  =  Vw*  -h  Nf, 

_tan/3 
tan2 


-  /3)     and      T  = 
C 


FIG.  314.     Wall  inclined  outward.     0  >  90°  +  8.     Surface  level. 


294 


GRAPHICS  AND   STRUCTURAL  DESIGN 


Va-cos2*  x  whj 
cos  a  +  v  cos2  a  —  cos2  <f>          2 


w  = 

E  = 


acts  parallel  to  line  of  fill. 

w  sin  /3  cos  (a  —  /?) 


2  cos2  ft  cos  a 


FIG.  315.    Wall  inclined  outward.     6  >  90° +5.     Surcharged. 


where 


tt-tarf  («•-£)  "(*•-*>, 


,. 
#2 


E  = 


/  cos0  y 


+  i/   2  cos  62 


and 


4  /s 
n=\ 
V 


sm  (0  +  5a)  sin  </> 


cos  §2 


tan/3  — 


tan  ^  = 


JF2, 

2  Nz  sin  52 
w  (H*  -  h*} 


tan2 


(«•-!) 


Q  =  E  cos  G*  —  /3);     T1 
sin  (/*  —  j8). 

NOTE.  —  The  quantity  Nz  sin  62  is  the 
friction  on  the  vertical  side  and  reduces 
the  weight  acting  upon  the  side  AB. 


FIG.  316.    Hopper  bin,  fill  level. 


BINS 


295 


cos  a.  —  Vcos2  a  —  cos2  <f>  xx  w  (hi2  — 
E\  =  cos  a  X  —  —  X 


cos  a  +  v  cos2  a  —  cos2  <f> 

sing  cos  (<*-/?)   .  «(ff_ 
cos2  /3  cos  a  2 


/  cos  <f>  \2     wh2  4  /sin 

#2=1- r  ,    where    n  =  V  ~ 

\  n  +  i  /    2  cos  62 


sin  (</>  —  a) 


cos  82  cos  a 


£  =      jVi2  +  W2  +  2  AW  sin  a. 
Ei  is  parallel  to  the  angle  of  the  fill. 


FIG.  317.   Hopper  bin,  surcharged, 
unsymmetrical . 


where      n  = 
W  = 


g_ 

cos2  0  cos  a  2 


£  cos  (jj.  -  /8)     and 
T  =  E  sin  On  -  /8). 


FIG.  318.    Hopper  bin,  surcharged,  sym- 
metrical. 


296 


GRAPHICS  AND  STRUCTURAL  DESIGN 


The  following  coefficients  of  friction  are  given  by  Mr.  Dull  in 
the  article  just  referred  to. 

COEFFICIENTS  OF  FRICTION   BETWEEN   MATERIALS 


Material. 

<f>  Degrees. 

<p  Degrees. 

Weight  per  cu. 
ft.,  pounds. 

Bituminous  coal  

7C 

18 

CO 

Anthracite  coal  

27 

16 

C2 

Sand     . 

•2A 

18 

oo 

Ashes  

4-O 

31 

40 

Where  the  material  runs  on  concrete  01  may  be  assumed  equal 


to 


=  coefficient  of  friction  of  the  material  upon  itself. 
1  =  coefficient  of  friction  of  the  material  upon  steel. 


Column  Load 

150000* 
FIG.  319.  FIG.  320. 

GRAPHICAL  DETERMINATION  OF  FORCES  ACTING  ON  BIN 

Bin  not  Surcharged.  — Assume  Fig.  319  to  be  a  bin  for  bitumi- 
nous coal.  The  coal  weighs  50  Ibs.  per  cu.  ft.  <£  =  35  degrees. 
The  natural  sine  of  35  degrees  is  0.574. 


BINS  297 

The  pressure  against  the  side  AB  and  the  plane  CI  may  be 
found  as  was  done  for  retaining  walls  with  Rankine's  formula. 
OnAB, 


=  5P. 


=  6 


2        i  +sin0        2  i  +0.574 

On  1C, 

1  79  1  —  sin  </>      50          9      0.426 

£2   =  -  rf   .  £  =  0_  x   2Q2   x  _         _ 

2  i  +  sm<£       2  !'574 

The  side  5C  is  acted  on  by  the  weight  of  the  superimposed  coal 
and  the  horizontal  pressure  E^.  It  is  easier  to  consider  the 
triangle  of  coal  C77,  and  then  having  found  the  pressure  on  CJ 
the  pressure  on  the  section  CB  may  be  scaled  from  the  diagram. 
The  weight  of  the  triangle  of  coal  CIJ  is  W  ••=  \  X  2O2  X  50  = 
10,000  lbs. 
This  resultant  force  acts  at  the  center  of  gravity  of  this  tri- 

/^*  7" 

angle  or  -  //  from  CI.     E%  acts  a  distance  CK  =  --  from  the 

3  3 

bottom  of  the  bin.  Combining  E%  and  W  gives  the  resultant 
R  =  10,300  lbs.  This  force  resolved  parallel  and  at  right  angles 
to  the  side  BC  gives  T  —  5200  lbs.  and  N  =  8900  lbs.  The 
normal  pressure  of  8900  lbs.  is  distributed  along  the  line  JC  as 
the  intercepts  in  the  triangle  LJC.  Putting  the  area  of  the 
triangle  LJC  =  8900  lbs.  and  solving  for  the  side  LC  gives  LC  = 
630  lbs.,  which  represents  the  pressure  per  square  foot  at  the 
point  C.  The  pressure  at  B  is  found,  by  scaling  the  intercept 
BMj  to  be  310  lbs. 

The  total  pressure  on  the  side  BC  is  the  area  of  the  trapezoid 
BCLM  and  equals  6580  lbs.  The  resultant  of  this  pressure  acts 
through  the  center  of  gravity  of  the  trapezoid.  The  forces 
acting  upon  the  points  of  the  bin  with  supports  every  12  ft.  are 
given  in  Fig.  320,  and  are  the  force  at  A  =  (-|--)  X  12  =  2700  lbs. 
The  horizontal  force  at  B  =  (f)  X675  X  12  =  5400  lbs. 

The  force  at  B  normal  to  the  side  BC  must  be  estimated  as 


298  GRAPHICS  AND   STRUCTURAL  DESIGN 

follows.  The  total  normal  pressure  on  BC  is  6580  Ibs.  CQ  = 
6.5  ft.  CB  =  14.1  ft.  The  resultant  at  B  is,  therefore, 

(6580X6,5}  =    Q      lbs 
14.1 

The  total  on  12  ft.  of  bin  is  3040  X  12  =  36,480  lbs.  The  balance 
of  the  load  represented  by  the  trapezoid  MBCL  acts  at  C  and  is 
(6580  X  12)  -  36,480  =  42,480  lbs. 

The  vertical  pressure  at  C  is  CD  X  DH  X^Xi2=5X2oX 
50  X  12  =  60,000  lbs. 

The  load  on  the  columns  due  to  the  fill  will  be  the  total  weight 
of  the  coal  in  the  bin.  The  area  of  the  bin  section  is  500  sq.  ft., 
making  the  load  in  12  ft.  500  X  12  X  50  =  300,000  lbs. 

The  trussed  bracing  will  be  carried  to  the  points  A,  B,  C,  E, 
F  and  G  of  the  bin  and  taking  these  forces  together  with  the 
total  column  load  the  several  forces  acting  in  the  frame  may  be 
determined. 

Surcharged  Bins.  —  A  bin  similar  to  the  last  one  but  sur- 
charged (Fig.  321)  may  have  the  forces  acting  upon  it  analyzed 
in  much  the  same  way.  First  find  the  pressure  upon  the  side 
AB]  this  is  done  graphically  in  Fig.  322.  The  diagram  gives 
p  =  8  ft.  and  y  =  8.5  ft.,  from  which  El  =  ^XwXpXy  = 
|  X  50  X  8  X  8.5  =  1700  lbs.  The  normal  pressure  is  1620  lbs. 

Now  finding  the  pressure  upon  IE,  the  axis  of  the  bin,  by  using 
the  graphical  method  and  assuming  <j>1  =  o,  we  have  p  =  y  = 
i6ft. 

Ez  =  0.5  X  w  X  p  X  y  =  0.5  X  50  X  16  X  16  =  6400  lbs. 

The  center  of  gravity  of  the  triangle  IJH  is  at  M  and  the 
weight  of  a  prism  of  coal  i  ft.  high  with  the  base  IJH  is 

(35.4  X  20.7  X  5°)  =  I8;300  lbs 

The  resultant  of  EZ  and  W  is  R  =  19,300  lbs.  This  is  assumed 
as  varying  uniformly  along  the  face  //.  The  pressure  at  7  is 

x  =  (19,300  X  2)  = 
19-3 


BINS 


299 


The  total  pressure  on  the  side  BC  is  that  represented  by  the 
trapezoid  QBCS  whose  altitude  scales  12.8  ft.;  the  area  then  is 
600  +  1550  g  _ 

2 


FIG.  321. 

This  acts  through  the  center  of  gravity  O  of  the  trapezoid 
which  has  been  located  graphically. 
The  normal  pressure  on  the  side 
BC  may  now  be  found  by  laying 
off  13,760  Ibs.  through  the  point 
0  and  resolving  it  into  its  com- 
ponents parallel  with  and  at  right 
angles  to  BC.  The  load  upon 
one-half  of  the  base  CD  is  7210  Ibs. 
The  forces,  due  to  the  fill,  acting  at  FlG 

the  points  of  the  bin  may  be  esti- 
mated as  in  the  preceding  problem  and  are  given  in  Fig.  323.  The 
distance  between  supports  has  been  taken  at  12  ft.    The  columns 


Column  Load 

232400* 


3oo 


GRAPHICS  AND   STRUCTURAL  DESIGN 


will  carry  the  total  weight  of  the  coal  in  the  bin  and  the  weight 
of  the  bin.  The  stresses  in  the  internal  framework  of  the  bin 
may  be  calculated  by  treating  these  external  loadings  as  is  done 
for  any  other  truss.  The  stress  diagram  may  be  drawn  or  the 
forces  calculated  by  any  of  the  methods  previously  given. 

SUSPENSION  BUNKERS 

If  the  side  plates  of  the  ordinary  hopper  bunker  are  permitted  to 
bulge  but  slightly  they  must  be  supported  at  frequent  intervals  by 
beams  or  other  structural  shapes.  This  added  weight  is  avoided 
in  the  use  of  suspension  bunkers.  These  bunkers  are  patented 
and  are  designed  with  the  idea  that  the  sides  resist  tension  only. 
To  fulfil  this  condition  theoretically  the  sides  would  assume  a 
shape  peculiar  to  each  possible  loading.  This  could  be  illustrated 
by  a  model  bin  whose  sides  were  a  purely  tension  piece  like  muslin 
or  duck.  However,  an  actual  bin  differs  greatly  from  the  above 
illustration  and  any  deflection  due  to  bending  in  the  sides  when 

the  bin  is  not  loaded  as  theoretically 
assumed  is  ordinarily  not  sufficient 
to   cause   trouble.      The   lined    bins 
will  be  stiffer  than  the  unlined  ones. 
The  theoretical  curve  of  a  bin  for 
any  assumed  loading  may  be  deter- 
mined  as   follows. 
In  Fig.  324,  given 


L 


n 


SUSPENSION 
BUNKERS. 

FIG.  324. 


desired    span 
sag,  the  load 


FIG.  325. 


the 

and 

in  the  left-hand  side 

of  the  bin  will  not 

vary  much  from  a 


triangle   ABC.      The  weight  of   the  material  in   the  portion 
ABC  of   the  bin  will   act  through  the  center  of  gravity  of 
this  triangle  ABC  and  lies  in  the  vertical  line  DG  or  W,  GF 
BF 

i        •  J  J 1 

being 

3 


BINS  301 

Considering  the  side  of  the  bin  BE  the  force  acting  at  B  and 
that  at  E  must  hold  W  in  equilibrium;  hence  they  must  pass 
through  a  common  point  D. 

The  line  of  action  of  W  is  known  as  is  also  the  line  of  action  of 
the  force  acting  at  E  which  must  be  horizontal  if  the  bin  is  sym- 
metrical and  symmetrically  loaded,  both  of  which  conditions  have 
been  assumed.  If  then  a  horizontal  line  is  drawn  through  E  it 
cuts  W  in  D  and  the  line  of  action  of  the  force  acting  at  B  must 
lie  in  the  line  BD  passing  through  the  point  D.  The  loading 
being  assumed  as  varying  as  the  intercepts  of  a  triangle  we  may 
represent  this  loading  by  the  triangle  FBC,  and  having  divided 
it  into  sections  of  equal  widths  the  weight  of  each  section  will 
be  proportional  to  its  center  ordinate;  these  have  been  shown 
dotted. 

These  lengths  from  1-2  to  5-6  have  been  laid  off  on  the  load 
line  of  the  force  polygon,  Fig.  325.  The  lines  of  action  of  the 
forces  acting  at  B  and  E  being  known  the  pole  0  is  readily  found 
by  drawing  Oi  parallel  to  DB  and  06  parallel  to  DE.  The  other 
strings  may  now  be  drawn  in  the  force  polygon.  If  we  complete 
the  equilibrium  polygon  in  Fig.  324  the  broken- line  side  will 
approximate  the  theoretical  side  of  the  bunker.  On  the  right- 
hand  side  the  line  of  the  bunker  has  been  drawn  as  it  is  com- 
monly made  in  practice.  The  portion  HI  is  straight  while  the 
lower  part  HE  is  an  arc  of  a  circle.  The  sides  of  this  bin 
approximate  an  equilateral  triangle.  The  sag  is  about  six-tenths 
of  the  span. 

For  a  bin  which  is  nearly  an  equilateral  triangle  the  following 
figures  will  be  approximately  true. 

Area  of  the  bin  section,  fill  level,  =  0.40  S2. 

Area  of  the  bin  section,  triangular  surcharge,  =  0.57  S1. 

Length  of  plate,  no  allowance  for  laps,  =  19.8  S  (inches). 

Force  in  plate  per  foot  of  length  at  B  or  /  =  0.60  W  (pounds). 

Force  in  plate  per  foot  of  length  at  £  =  0.30  W  (pounds). 

Here  5  =  the  span  in  feet. 

W  =  the  load  in  the  bin  for  one  foot  of  its  length,  pounds. 


GRAPHICS  AND   STRUCTURAL  DESIGN 


FIG.  326. 


In  the  case  of  the  suspension  bunker  with  vertical  sides  above 

the  points  of  suspension,  B  and  7  of 
the  bin,  Fig.  326;  the  cross  section 
of  the  bunker  will  be  increased  by 
the  area  BCIJKL  equal  to  S  X  h, 
and  the  load  will  be  increased  pro- 
portionally. In  this  case  the  weight 
of  the  material  in  one-half  of  the  bin 
will  act  through  the  center  of  gravity 
of  the  section  KLBDEK.  This  will 
be  a  greater  distance  from  the  cen- 
ter line  EC  than  was  the  case  in  the 
preceding  problem,  and  will  make 

the  bin  somewhat  wider  below  BI  than  at  corresponding  points 

in  the  former  bin. 

W 

In  either  problem  if  --  is  known  the  forces  acting  at  B  and  E 

may  be  found  by  drawing  the  triangle  B1E1D1  at  the  side  of  the 
bin  as  in  Fig.  326. 

One  form  of  Suspension  Bunker  with  column  supports  for 
boiler  rooms  is  shown  in  Fig.  327.  This  bunker  is  intended  to 
hold  6600  Ibs.  per  ft.  of  length.  Assume  coal  as  weighing  50  Ibs. 
per  cu.  ft.,  and  the  sag  of  the  bunker  as  60  per  cent  of  the  span. 
The  bunker  will  be  assumed  as  carrying  the  required  load  when 
surcharged,  and  the  angle  of  repose  of  the  coal  will  be  taken  30 
degrees.  Determine  the  bin  area  by  using  the  formula,  A  = 
0.57  S2.  The  required  area  is  ^-f^-  =  132  sq.  ft.  From  this 
132  =  0.57  S2  and  S  =  15.3  ft.  If  the  bin  is  made  16  ft.  wide 
the  sag  will  be  approximately  10  ft.  Spacing  the  columns  20  ft. 
center  to  center  makes  the  load  of  coal  upon  each  column 

- =  66,000  Ibs.     Adding  4000  Ibs.  to  this  to  cover 

2 

the  weight  of  metal,  exclusive  of  columns  and  under  bracing, 
makes  a  total  of  70,000  Ibs.  at  each  column.  This  load  laid  off 
in  Fig.  328  gives  the  stress  in  the  side  of  the  bunker  per  foot  of 


BINS  303 

length  as  -        -  =  4000  Ibs.     The  unit  stress  per  square  inch 


20 


in  a  plate  -  in.  thick  is 


4000 


=  1330  Ibs. 


-  in.  tmcK  is  / 

4  (0.25  X  12) 

The  main  columns  are  subjected  to  a  bending  moment  of 
80,000  [(16  —  6)  —  6]  =  320,000  in.  Ibs. 


FIG.  328. 


Trying  two  i2-in.  channels  weighing  20.5  Ibs.  per  ft.,  we  have 

Fiber  stress  due  to  the  bending  moment  is  ,       ' — ^-  =  7,400  Ibs.  per  sq.  in. 


,        — 

\2  X  21. 


Fiber  stress  due  to  direct  loading  is 


80,000 
(2  X  6.02) 

Combined  stress 


=  6,650  Ibs.  per  sq.  in. 


=  14,050  Ibs.  per  sq.  in. 

The  direct  load  upon  the  column  being  6650  Ibs.  per  sq.  in.  the 


304  GRAPHICS  AND   STRUCTURAL  DESIGN 

radius  of  gyration  for  an  assumed  length  of  n  ft.  may  be  found 
from  the  straight-line  formula, 

/  =  16,000  —  70  M  =  16,000  —  70 

70  X  1^2 

or  r  =  y—f—  —  =  0.99. 

(16,000—  6650) 

Two  i2-inch  channels  are  to  be  used  back  to  back.  The 
radius  of  gyration  of  a  1 2-inch  channel  referred  to  an  axis  at  the 
back  of  the  web  is 


_  Ah 

= 


*  _  v/3.9o  +  (6.02  X  o.yo2)       T    , 
V  ~~" 


It  therefore  follows  that  two  i2-in.  channels  at  20.5  Ibs.  per  ft. 
either  riveted  back  to  back  or  separated  any  distance  will  carry 
the  load  of  80,000  Ibs.  applied  axially. 

The  horizontal  strut  under  the  bin  carries  a  load  of  40,000  Ibs. 
Its  length  is  84  ins.  Trying  a  lo-in.  channel  at  15  Ibs.  per  ft., 
whose  radius  of  gyration  about  an  axis  parallel  to  the  web  is  0.70, 

we  find  -  =  —  *-  =  1  20.     The  allowable  stress  is  /  =  16,000  — 
r      0.70 

(70  X  120)  =  7600  Ibs.  per  sq.  in.     The  required  area  then  is 

-  =  5.26   sq.  ins.     The   strut  across   the   top  of  the  bin 
7600 

carries  the  conveyor,  its  truss  and  the  conveyor  loading  and  in 
addition  to  these  the  direct  compression  due  to  the  bin  loading. 
Assuming  the  load  equivalent  to  a  central  load  of  3500  Ibs.  the 

bending  is  M  =  -    -  =  3500  X  16  X  —  =  168,000  in.  Ibs.     Trying 
4  4 

a  i2-in.  I  beam  at  25  Ibs.  per  ft.,  with  an  -  value  of  24.6  and  an 

c 

area  of  7.34  sq.  ins.;  we  have 

Fiber  stress  due  to  bending  moment,  —  '——  =  6800  Ibs.  per 

24.0 

sq.  in. 


BINS  305 

Fiber  stress  due  to  direct  loading,  — =  5450  Ibs.  per  sq.  in. 

7-34 

The  combined  fiber  stress  is  6800  +  5450  =  12,250  Ibs.  per  sq. 
in.  Considering  that  this  beam  is  held  laterally  at  its  center 
by  the  conveyor  truss  this  fiber  stress  would  seem  permissible. 
The  braces  occur  every  10  ft. 

The  side  of  the  bin  between  columns  is  carried  by  a  girder. 
The  total  load  is  80,000  Ibs.  or  4000  Ibs.  per  ft.  The  bending 
moment  then  is 

Wl  12 

M  =  -  -  =  80,000  X  20  X  --  =  2,400,000  in.  Ibs. 
8  o 

Assuming  the  distance  back  to  back  of  flange  angles  as  3  ft., 
the  distance  between  the  centers  of  gravity  of  the  two  flanges 
will  be  about  34  ins. 

M  =  A  Xf  X  h  =  2,400,000  =  A  X  16,000  X  34 
or  A  =  4.42  sq.  ins. 

Two  5  X  3^  X  TVm-  angles  having  a  gross  area  of  5.12  sq.  ins. 
and  a  net  section,  allowing  for  one  rivet,  of  4.52  sq.  ins.  furnish 
just  a  little  more  than  the  area  required.  No  allowance  has  here 
been  made  for  the  web  acting  as  flange.  This  gives  additional 
security. 

Figure  329  shows  a  bin  designed  for  bituminous  coal.  The 
weight  of  the  coal  was  taken  50  Ibs.  per  cu.  ft.,  the  natural 
slope  was  taken  30  degrees,  while  the  angle  of  friction  between 
the  sides  and  the  coal  was  assumed  as  18  degrees.  The  diagram 
gives  p  =  1 6  ft.  and  y  =  17  ft. 

E  =  %Xw- p. y  =  %X  $0X16X17  =  6800  Ibs. 

The  horizontal  component  of  this  force  is  6500  Ibs.  The  hori- 
zontal pressure  per  square  foot  on  the  sides  of  the  bin  at  the 
bottom  is  x  =  6500  X  -$$  =  433  Ibs. 

The  sides  are  supported  by  I  beams  spaced  6  ft.  center  to 
center  and  horizontal  rods  are  run  across  the  bin  at  6-ft.  intervals 
and  placed  10  ft.  above  the  bottom  of  the  bin.  The  stresses  were 


306  GRAPHICS  AND   STRUCTURAL  DESIGN 

determined  graphically  as  outlined  for  fixed  and  continuous 
beams.  The  force  in  the  tie  rods  was  34,200  Ibs.  and  the  maxi- 
mum bending  moment  on  the  I  beams  was  found  to  be  507,000 
in.  Ibs.  Bolts  if  ins.  in  diameter,  having  an  area  of  2.05  sq.  ins. 

at  the  root  of  the  threads,  would  have  a  unit  stress  of  34>20°  = 

2.05 

16,500  Ibs.  per  sq.  in.  and  are  satisfactory. 
Allowing  18,000  Ibs.  per  sq.  in.  fiber  stress  in  the  beam  would 

require  an -value  of   -  =  —r  =  S0?'000  =  2g  2      This  Would 
e  e        f         18,000 

require  a  lo-in.  I  beam  at  35  Ibs.  per  ft.  or  a  i2-in.  I  beam  at 
31.5  Ibs.  per  ft.  This  bin  was  tied  across  the  top  by  the  roof 
framing  and  was  braced  transversely  at  intervals  against  wind 
pressure. 

BIN  DESIGN 

Figure  330  illustrates  a  bin  for  coal  storage  over  gas  producers. 
The  upper  portion  of  the  bin  was  a  square  of  13  ft.  side.  The 
lower  part  was  a  curve  with  a  g-ft.  sag.  The  coal  is  assumed  as 
weighing  50  Ibs.  per  cu.  ft.  and  its  angle  of  repose  has  been  taken 
35  degrees. 

The  angle  of  friction  between  the  coal  and  the  sides  of  the 
bin  has  been  taken  18  degrees.  Owing  to  the  large  amount  of 
coal  carried  in  the  square  portion  of  the  bin  the  loading  cannot 
be  represented  as  a  triangle;  therefore  an  approximate  outline 
of  the  bin  has  been  assumed  in  Fig.  330  and  the  actual  curve 
estimated  from  this.  The  left  side  of  the  bin  has  been  divided 
into  four  strips  of  equal  widths;  the  weight  in  each  strip  per  foot 
of  length  of  bin  will  be  proportional  to  the  middle  ordinate, 
shown  in  full  lines  for  each  section.  In  the  force  polygon,  Fig. 
331,  the  forces  may  be  represented  by  these  middle  ordinates  or 
by  a  constant  percentage  of  them.  In  the  figure  the  scale  has 
been  taken  as  one-fourth  of  these  lines.  The  forces  AlBl,  BlCl, 
etc.,  have  been  laid  off  on  the  right-hand  side  of  the  bin  equal  to 
and  symmetrically  placed  with  those  on  the  left-hand  side.  These 


BINS 


307 


forces  have  been  laid  off  on  the  load  line  of  the  force  polygon, 
Fig.  331,  and  the  vector  diagram,  on  its  left-hand  side,  has  been 


used  to  locate  the  center  of  gravity  of  the  load  in  the  right-hand 
side  of  the  bin.  The  center  of  gravity  is  found  in  the  inter- 
section of  the  lines  i  and  5  in  Fig.  330.  R,  the  total  weight  of 
the  coal  on  this  side  of  the  bin,  acts  through  this  point  of  inter- 
section. The  force  in  the  extreme  low  point  of  the  bin  will  be 
horizontal,  and  this  together  with  the  force  in  the  bin  side  acting 
through  c  must  hold  R  in  equilibrium,  hence  these  three  forces 
must  pass  through  a  common  point  b ;  this  gives  the  direction  of 
be,  which  is  the  theoretical  direction  of  the  side  of  the  bin  at  the 
point  c. 

The  lines  for  the  other  side  of  the  bin  have  been  made  similar 
to  those  for  the  right-hand  side.  The  vector  polygon  drawn  on 
the  right-hand  side  of  the  load  line,  Fig.  331,  has  been  used  in 
drawing  the  dotted  lines  in  Fig.  330,  thus  showing  the  agreement 
between  the  actual  and  the  theoretical  lines  of  the  bin. 

Before  the  stresses  can  be  read  in  Fig.  331,  the  scale  must  be 
found. 


GRAPHICS  AND   STRUCTURAL  DESIGN 


The  mean  ordinate  in  one  side  of  the  bin  is  21  ft.  The  bin 
being  13  ft.  wide  makes  the  volume  of  the  bin  section  for  a  length 
of  i  ft.  =  21  X  13  =  273  cu.  ft.  This  gives  273  X  50  =  13,650 
Ibs.  of  coal  per  ft.  of  bin.  The  load  line  AE  =  6825  Ibs.  This 
makes  the  scale  $-&££-  =  1700  Ibs.  per  in.  (approximately). 

The  force  A0l  =  4.3  X  1700  =  7300  Ibs. 

The  force  E0l  =  1.6  X  1700  =  2720  Ibs. 

The  bin  plates  will  be  made  f$  in.  in  thickness  to  allow  for 
wear  and  deterioration,  and  will  have  3  X  f-in.  wearing  strips 
spaced  about  12  ins.  center  to  center  on  the  curved  portion  of 
the  bin  sheets  and  running  the  length  of  the  bin.  The  distance 
center  to  center  of  columns  will  be  assumed  as  18  ft.  Trans- 
verse braces  will  be  placed  at  the  columns  and  every  9  ft. 


FIG.  332. 


FIG.  333. 


The  pressure  against  the  vertical  sides  of  the  bin  is  determined 
in  Fig.  332. 

E  =  %  Xw  •  p  -y  =  |  X  50  X  10.5  X  n  =  2890  Ibs. 

It  should  be  noted  that  for  so  narrow  a  bin  this  does  not  give 
the  accurate  pressure  but  it  is  the  limiting  pressure  to  which 
this  force  on  the  side  of  the  bin  could  extend.  A  closer  approxi- 
mation might  have  been  made  by  assuming  a  bin,  say,  16  ft. 
high  instead  of  13  ft.  but  with  a  horizontal  fill.  The  horizontal 
component  of  this  force  of  2890  Ibs.  is  2700  Ibs.  Two-thirds  of 
this,  or  1800  Ibs.,  act  at  the  lower  points  e  and  c,  while  the  other 
third  act  at  the  upper  points  /  and  g.  These  are  each  900  Ibs. 
These  forces  act  in  the  opposite  direction  to  the  pull  from  the 
suspended  portion  of  the  bin. 


BINS  309 

Had  the  bin  been  level,  substitution  in  the  formula  for  the 
pressure  on  the  side  of  the  bin  would  have  given  1060  Ibs.  or  700 
Ibs.  at  the  points  e  and  c;  the  net  pressure  here  then  would  be 
2700  —  700  =  2000  Ibs.  per  ft.  of  length  of  bin. 

The  compression  in  the  lower  member  of  the  brace  due  to  the 
total  load  between  the  columns  is  2000  X  18  =  36,000  Ibs.  If 

the  value  of  -  in  these  braces  is  to  be  limited  to  120  then  r  =  13 

X  iV\  =  i-3;  and  this  requires  two  6  X  3^-in.  angles  having 
their  long  legs  placed  back  to  back.  Before  the  weight  is  selected 
the  possibility  of  additional  loading  from  wind  should  be  con- 
sidered. 

In  exposed  places  this  might  reach  30  Ibs.  per  sq.  ft.  of  vertical 
projection.  Along  the  bin  at  the  point  e  it  would  equal  (6.5  +  9) 
X  30  =  465.  This  makes  the  total  load  at  the  column  (465  X  18) 
+  36,000  =  44,37°  lbs- 

The  unit  stress  on  the  angle  is 


f 


16,000  —  f  70  X  - J  =  7600  lbs.  per  sq.  in. 


The  net  area  of  the  angles  is       ^*     =  5.83  sq.  ins. 

7600 

The  minimum  angles  should  therefore  prove  ample,  unless 
larger  angles  are  used  to  provide  for  corrosion  and  possible  injury 
from  the  coal  passing  over  them. 

The  channel  at  e  is  acted  upon  by  the  wind  and  the  pull  from 
the  side  of  the  bin.  Considering  first  the  wind  load,  it  will  act  as 
a  uniform  load  on  the  span  of  18  ft.  Its  deflection  under  this 
load  will  put  a  central  load  on  the  channel  at  c.  If  P  is  the  load 
transferred  to  the  channel  at  c  from  the  channel  at  e,  the  de- 

r  /W  .  /3v 

flection  due  to  the  wind  load  W  is  -  -  X  ( — — - ) ,  while  the  de- 

384      \h*lj 

i         iP  •  13\ 

flection  due  to  the  central  load  is  --  X  [  — — ;.)•      Since   the 

48       \Jb  •  L  / 

deflections  of  the  two  channels  are  equal, 


310  GRAPHICS  AND   STRUCTURAL  DESIGN 


_       __     __ 
384  El  '    48  El  ~  48  El 

and  =  L     or     P  =  —  W  =  0.31  W. 

384       24  384 

The  maximum  bending  moment  may  now  be  found  by  adding 
algebraically  the  bending  moments  due  to  these  several  loads. 
The  moment  due  to  the  wind  load  of  465  Ibs.  per  ft.  is 

M  =  W  -  -  =  465  X  18  X  18  X  -V2-  =  226,000  in.  Ibs. 
8 

The  bending  moment  due  to  the  central  load  is 

M  =  P  .-  =  0.31  X  465  X  18  X  18  X  -V"  =  140,000  in.  Ibs. 

4 

The  bending  moment  due  to  the  uniform  load  of  2000  Ibs.  per  ft. 
on  the  p-ft.  span,  that  is  between  the  braces,  is 

M  =W--  =  2000  X  9  X  9  X  -V2-  =  243,000  in.  Ibs. 
8 

The  resultant  bending  moments  are  given  by  Fig.  333,  and  are 
the  intercepts  between  the  curve  abc  and  the  lower  curves,  aedjc. 
The  curve  abc  gives  the  moments  due  to  the  wind  load  acting 
on  the  i8-ft.  span  and  measured  from  the  base  line  ac.  The 
moments  due  to  the  central  load  on  the  i8-ft.  span  are  then 
deducted  by  laying  off  the  triangle  adc.  The  bending  moments 
due  to  the  uniform  load  on  the  9-ft.  span  are  then  plotted  below 
ad  and  dc,  thus  giving  the  algebraic  sum  of  these  bending  mo- 
ments. The  maximum  moment  scales  340,000  in.  Ibs.  and  would 

,  ,  /      ,        ,IM      340,000 

require  a  channel  having  an  -  value  of  -  =  —  r  =  ^°-    —  =  2I-3> 

e  e       f        16,000 

which  calls  for  a  i2-in.  channel  at  20.5  Ibs.  per  ft.     The  fiber 

stress  in  this  channel  would  then  be  ^—  -  -  =  15,900  Ibs.  per 

21.4 

sq.  in. 

In  a  similar  way  the  maximum  bending  on  the  upper  channels 
is  found  to  be  approximately  139,000  in.  Ibs.  and  requires  a 

section  modulus  of  -~  -  =8.7. 
16,000 


BINS  311 

The  sides  act  as  girders  to  transfer  the  loads  to  the  columns. 
The  uniform  load  due  to  nil  is  13,650  Ibs.  per  ft.,  and  assuming 
the  weight  of  the  bin  as  approximately  1000  Ibs.  per  ft.  makes 
the  total  load  per  foot  14,650  Ibs.  The  load  per  foot  on  each 

girder  is  *4'  5°  =  7325  Ibs. 

The  bending  moment  on  the  girder  is 

M  =  ™~l  =  7325  x  18  X  v  =  296,660  ft.  Ibs. 

o  o 

If  the  girder  is  assumed  as  13  ft.  deep  the  flange  force  is 
206,660 


22,800  Ibs. 
J3 

When  it  is  considered  that  in  the  lower  channel  the  splice  plate 
to  which  the  curved  portion  of  the  bin  is  attached  assists  in 
carrying  this  stress  it  will  be  seen  that  this  force  of  22,800  Ibs. 
will  add  very  little  to  the  fiber  stress  in  the  i2-in.  channel. 

If  lo-in.  channels  at  15  Ibs.  per  ft.  are  tried  for  the  upper 
channels  the  fiber  stress  due  to  the  bending  is 

.      M-e      139,000 

/  =  — —  =  -         -  =  10,400  Ibs.  per  sq.  in. 
*  I3-4 

The  fiber  stress  due  to  the  flange  force  of  22,800  Ibs.  is 

22,800 

-  =  5100. 
4.46 

The  combined  fiber  stress  is  10,400  -f  5100  =  15,500  Ibs.  per 
sq.  in. 

The  stresses  may  be  increased  by  wind  acting  on  the  ends 
of  the  bin  or  building  and  also  by  thrust  from  the  traveling 
crane,  or  the  forces  may  be  provided  for  by  additional  bracing. 

The  stiffeners  on  the  sides  ef  and  cq  of  the  bin  carry  a  triangular 
load  of  2700  Ibs.  per  ft.  of  length  of  the  bin;  hence  if  the  stiffeners 
are  spaced  2  ft.  3  ins.  center  to  center,  their  total  load  is  2700  X 
2.25  =  6070  Ibs.  The  moment  on  a  beam  due  to  such  a  load  is 
M  =  0.128  XP-/;M=  o.i28X  6o7oX  13  X  12=  121,000  in.  Ibs. 


312 


GRAPHICS   AND   STRUCTURAL   DESIGN 


=  7-6, 


The  section  modulus  required  is 

/  _  M      121,000 
e       f        16,000 
calling  for  8-in.  channels  at  uj  Ibs.  per  ft. 

The  maximum  stress  in  the  -^g-in.  plates  along  the  side  of  the 
bin  will  occur  at  e.  Since  the  load  of  2700  Ibs.  is  assumed  as 
distributed  as  a  triangle  of  pressure,  it  follows  that  if  pe  is  the 

pressure  per  square  foot  at  e,  then  2700=  -3-2s  and  ^  =  415  Ibs. 
per  sq.  ft. 


lo"Channel®15lfper  foot 


3-6x4x^1? 


\ 


,10'p<il.V*  / 


/ 


12"Channel  g  20J£  ^  per  foot 


FIG.  334. 


FIG.  335. 


The  pressure  per  square  inch  is  {J|  =  2.88  Ibs. 

27,2 

The  maximum  fiber  stress  is  fm  =  -  0  —      —  X 

2     a2  +  b2 


fm=-2Xi.oX 


X 


2.88 


=  10,600  Ibs.  per  sq.  in. 


A  cross  section  and  end  view  of  this  bin  is  shown  in  Fig.  334, 
while  the  elevation  is  given  in  Fig.  335.  The  diagonal  bracing 
shown  in  the  half  section  has  not  been  drawn  in  the  elevation. 


BINS  313 

It  would  be  located  at  the  columns  and  at  the  center  line  midway 
between  them.  In  general  only  the  bin  proper  has  been  shown, 
as  the  design  of  the  wind  bracing  and  columns  would  have  been 
complicated  by  the  character  of  the  building  and  it  was  not 
desired  to  consider  this  design  here. 


CHAPTER   XIX 
SHOP   FLOORS 

THE  purpose  of  a  floor  is  primarily  to  carry  such  loads  as  men, 
machines  and  materials.  In  addition  they  should  preferably 
be  poor  conductors  of  heat  as  this  adds  greatly  to  the  comfort 
of  men  standing  on  them. 

They  should  also  be.  durable,  resisting  wear  from  the  traveling 
of  men  and  the  transportation  of  goods  upon  them.  They  should 
furthermore  be  impervious  to  moisture.  In  special  cases  addi- 
tional qualities  may  be  desirable;  thus  in  many  floors  it  is  essential 
that  they  shall  be  nearly  dustless,  while  in  other  cases  a  somewhat 
soft  surface  may  be  desired. 

Earth  and  cinder  make  the  cheapest  floor.  Owing  to  dust  they 
are  only  adapted  to  places  like  forge-shops,  where  hot  metals 
placed  upon  them  or  heavy  pieces  dropped  on  them  would  only 
injure  other  floors.  The  soil  below  the  floor  level  should  be 
drained  if  damp  and  the  floor  clay  or  cinder  then  placed  in  layers 
upon  a  rolled  bed;  each  layer  should  be  thoroughly  tamped  or 
rolled.  Where  the  moisture  is  excessive  this  floor  may  be  made 
impervious  by  a  layer  of  tar  concrete  placed  from  1.5  to  2  ft. 
below  the  floor  surface. 

Cement  Concrete  Floors.  —  In  a  floor  of  this  type  it  is  necessary 
to  have  a  substantial  foundation ;  its  depth  will  vary  with  the 
character  of  the  soil  from  i  to  2  ft.  The  soil  having  been  ex- 
cavated to  the  desired  depth  the  surface  should  be  well  tamped 
or  rolled.  A  layer  of  broken  stone,  cinder  or  gravel  6  ins.  to 
10  ins.  thick  is  placed  upon  this  and  thoroughly  tamped.  In 
some  floors  a  layer  of  from  3  ins.  to  6  ins.  of  finely  crushed  stone 
is  placed  upon  this  and  well  rammed,  this  being  in  turn  covered 
w?th  2  ins.  to  4  ins.  of  concrete  upon  which  a  wearing  coat  of 


SHOP  .FLOORS  315 

cement  mortar  from  \  in.  to  i  in.  thick  is  placed.  In  other 
designs  the  concrete  is  placed  upon  the  first  foundation  layer. 
This  concrete  is  3  ins.  to  4  ins.  thick  and  upon  it  a  wearing  coat 
of  i  in.  of  cement  mortar  is  placed. 

The  concrete  is  usually  a  mixture  of  i  part  cement,  2  to  3  parts 
of  sand,  and  5  to  6  parts  of  stone  or  gravel.  The  cement-mortar 
is  i  part  cement  to  2  parts  of  sand  or  may  be  richer  in  cement. 
Where  the  soil  affords  a  sufficiently  good  foundation  5  ins.  to 
6  ins.  of  concrete  placed  immediately  upon  it  may  prove  satis- 
factory. 

Tar  Concrete  Floors.  —  Where  a  wooden  wearing  surface  is 
desired  such  surface  may  be  laid  upon  the  concrete  base  just 


FIG.  336. 

described,  in  which  case  the  surface  of  the  concrete  under  the 
wood  should  be  coated  with  coal  tar  or  asphalt;  otherwise  the 
wood  deteriorates  from  dry  rot. 

Another  plan  (Fig.  336)  makes  the  foundation  of  a  coal  tar 
or  an  asphalt  concrete.  After  the  top  soil  is  removed  the  surface 
is  leveled  and  rolled  and  a  layer  of  from  4  ins.  to  6  ins.  of  broken 
stone  or  gravel  thoroughly  mixed  with  tar  is  placed  and  well 
rolled.  A  i-in.  layer  of  sand  thoroughly  saturated  with  tar  is 
put  on  this  concrete  and  carefully  rolled.  While  the  tar  is  still 
hot  hemlock  planking  is  put  on  the  sand  and  pushed  into  place. 
The  wearing  surface,  which  is  preferably  maple  strips,  tongued 
and  grooved,  is  placed  at  right  angles  to  and  upon  this  hemlock. 
In  some  cases  the  foundation  is  made  like  that  described  for  a 
concrete  floor.  The  design  may  also  be  modified  by  using  hem- 
lock sleepers  instead  of  the  planking. 


316  GRAPHICS  AND  STRUCTURAL  DESIGN 

In  this  case  the  hemlock  strips  are  placed  on  about  3-ft.  to 
4-ft.  centers,  being  imbedded  in  the  sand  and  tar,  and  the  tongued 
and  grooved  maple  flooring  is  then  nailed  to  the  sleepers. 

Wooden  Block  Floors.  —  Wooden  blocks  set  on  end  make  an 
excellent  floor.  These  blocks  are  about  4  ins.  to  5  ins.  thick  and 
may  be  either  cedar,  maple,  beech,  oak  or  pine.  When  not 
treated  with  a  preservative  they  should  be  set  in  coal  tar.  Also 
when  not  treated  the  blocks  should  be  set  with  spaces  between 
them  to  allow  for  swelling  due  to  any  moisture  they  may  absorb. 
These  spaces  between  the  blocks  should  be  filled  with  tar  and 
sand.  These  blocks  may  be  placed  upon  the  tar-concrete 
foundation  just  described;  see  Fig.  336. 

Asphalt  Floors.  —  Asphalt  makes  a  satisfactory  floor  in  many 
places,  but  it  is  not  suited  to  machine  shops  or  places  where  much 
oil  is  likely  to  reach  it.  Asphalt  gives  a  soft  surface,  is  water- 
proof, dustless  and  a  poor  conductor  of  heat.  It  therefore  makes 
a  very  comfortable  floor  for  the  men  to  stand  upon.  It  is 
easily  dented  but  the  dents  will  gradually  smooth  out  owing  to 
the  tendency  of  the  asphalt  to  flow  under  pressure. 

The  foundation  is  similar  to  that  for  a  concrete  wearing  surface. 
The  asphalt  is  generally  made  an  inch  thick.  A  wash  of  molten 
tar  and  asphalt  should  be  put  on  the  surface  of  the  concrete 
before  placing  the  asphalt.  Where  asphalt  is  placed  upon  a  con- 
crete slab  it  forms  an  additional  dead  load  and  should  not  be 
assumed  as  resisting  any  of  the  compression  in  the  slab. 

Brick  Floors.  —  Brick  floors  are  specially  adapted  to  places 
where  a  floor  is  liable  to  local  injury,  as  no  other  floor  is  so  readily 
repaired,  it  being  possible  to  remove  and  replace  a  few  bricks 
without  disturbing  a  large  section  of  the  floor.  It  has  there- 
fore found  favor  in  railroad  buildings  such  as  round  houses.  For 
brick  paving  the  foundation  consists  of  from  4  to  6  ins.  of  cinder 
placed  in  layers  and  tamped  or  rolled  upon  the  excavated  surface 
which  had  also  been  previously  compacted.  A  layer  of  sand 
from  i  to  2  ins.  thick  is  placed  and  rolled  upon  the  cinder  and 
then  the  brick  are  laid  upon  the  sand.  The  bricks  should  be 


SHOP  FLOORS  317 

hard,  homogeneous  and  impervious  to  water.  They  should  be 
set  on  edge  and  joints  should  be  broken  by  a  lap  of  at  least 
3  ins.  None  but  whole  bricks  should  be  used  except  when  nec- 
essary as  fillers.  After  laying,  sand  should  be  thoroughly  worked 
into  the  spaces  between  the  bricks. 

To  make  this  floor  waterproof  the  brick  may  be  grouted  with 
tar  and  pitch,  after  which  the  sand  may  be  worked  into  the 
spaces  as  before. 

Wooden  Floors.  —  These  are  commonly  made  by  embedding 
stringers,  timber  strips,  in  a  foundation  similar  to  those  described 
for  concrete  floors  and  then  nailing  the  wearing  surface  of  the 
floor  on  the  stringers. 

One  wooden  floor  had  a  foundation  of  8  ins.  of  cinders,  the 
stringers  were  spaced  3  ft.  centers  and  3 -in.  planking  was  placed 
upon  these. 

The  stringers  and  under  surface  of  the  planks  were  coated 
with  lime  to  preserve  them.  Present  practice  prefers  tar  to 
lime  to  preserve  wood.  In  other  cases  the  foundation  is  a  layer 
of  concrete  in  which  the  stringers  are  embedded;  the  under  sur- 
face of  the  wood  should  be  coated  with  tar  as  before. 

Ground  Floors  in  General.  —  Where  much  water  is  used  on 
floors  they  should  be  given  sufficient  pitch  to  properly  drain  the 
water  and  suitable  drains  or  gutters  must  be  provided. 

Where  heavy  trucking  is  done  tracks  must  be  laid.  These 
may  be  either  tracks  sunk  in  the  floor  or  flat  iron  plates  placed 
in  the  floor  level  with  the  wearing  surface  of  the  floor.  The 
former  method  is  objectionable  on  account  of  breaking  into  the 
surface  of  the  floor.  The  flat  plates  should  be  corrugated  to 
prevent  their  becoming  too  smooth. 

UPPER  FLOORS 

Floors  above  ground  level  usually  consist  of  two  parts,  one 
affording  the  necessary  strength  to  sustain  the  floor  loads,  the 
other  supplying  the  wearing  face.  The  first  part  may  be  wood, 
steel,  concrete,  brick  or  tile,  or  may  be  a  combination  of  some 


GRAPHICS  AND   STRUCTURAL  DESIGN 


of  these.     The  wearing  surface  may  be  any  of  the  wearing 
materials  used  upon  the  ground  floors. 

Wooden  Floors.  —  A  wooden  floor  is  commonly  carried  upon 
wooden  joists  or  steel  floor  beams.  The  depth  of  the  floor 
timbers  furnishing  the  strength  will  vary  with  the  distance 

u<r\  Tongued  and  Grooved  Maple 


1 

1 

falling  Strip 

• 

<r— 

T 

FIG.  337. 

center  to  center  of  the  joists  or  floor  beams  and  with  the  floor 
loads  and  will  range  from  2  or  3-in.  planking  as  in  Fig.  337  to 
2  X  8-in.  timbers  placed  on  edge  and  nailed  together  as  in  Fig. 
338.  These  planks  should  be  continuous  over  at  least  2  supports. 
The  joints  should  be  staggered,  that  is  not  broken  continuously 


FlG.  338. 

on  the  same  line.  The  wearing  surface  will  usually  be  i  to 
i.5~in.  maple  or  other  hardwood  flooring.  Frequently  3  thick- 
nesses of  rosin-sized  paper  are  placed  between  the  wearing  and 
the  supporting  wood.  Where  the  timbers  are  on  edge  a  layer 
(^  in.)  of  tar  cement  (tar  and  sand)  may  be  placed  between 
the  two  sections. 


SHOP  FLOORS 


319 


Steel  Floors.  —  The  forms  shown  in  Figs.  339  and  340  run  to 
considerable  weight  and  are  suited  to  heavy  loads.  The  fill 
usually  consists  of  a  cinder  concrete  which  need  not  be  richer  than 
i  part  cement,  2  parts  sand  and  6  parts  cinder.  This  is  carried 


A A 


A 


J 

T 

it"   ^ 



FIG.  339. 

FIG.  340. 

a  few  inches  above  the  top  of  the  trough  section.     Sleepers  are 
embedded  in  this  fill  and  the  wearing  boards  nailed  to  them. 

There  are  other  trough  sections  made  of  much  lighter  metal, 
gauges  No.  16  to  No.  24.  These  sections  are  shown  in  Figs.  341 
to  343.  These  are  made  in  both  black  and  galvanized  steel. 
The  exposed  side  requires  painting. 


FIG.  341. 


FIG.  342. 


FIG.  343. 


Brick  Arch.  —  Another  heavy  floor  construction  is  shown  in 
Fig.  344.  It  consists  of  a  common  brick  arch  having  a  rise  of 
preferably  not  less  than  one-eighth  of  the  span. 

The  tie  rods  should  be  placed  at  frequent  intervals  along  the 
beams,  especially  along  the  outside  beams,  to  prevent  too  great 
stress  in  the  beams  from  lateral  pressure.  The  horizontal  thrust 
or  pressure  in  pounds  per  lineal  foot  of  arch  is  given  in  the  Pen- 
coyd  handbook  as 


320 


GRAPHICS  AND  STRUCTURAL  DESIGN 


W  =  load  in  pounds  per  square  foot  on  the  floor. 
5  =  span  of  arch  in  feet. 
R  =  rise  of  arch  in  inches. 


FIG.  344. 

The  fill  above  the  arch  may  be  a  lean  cinder  concrete,  say, 
1-2-6. 

The  wearing  face  may  be  any  of  those  previously  mentioned 
but  is  commonly  maple. 


: : — " — ~^~ — : r 


FIG.  345. 

The  arch  instead  of  being  made  of  common  brick  may  be  any 
of  the  several  types  of  hollow  clay  tile  (Fig.  345),  or  may  be  a 
concrete  arch  placed  on  corrugated  steel,  expanded  metal  or 


FIG.  346. 

wire  cloth;  see  Fig.  346.     These  floors  are  made  suited  to  lighter 
floor  loads  than  the  first-described  arch. 

The  principal  purpose  of  the  corrugated  steel,  expanded  metal 
or  wire  cloth  is  to  carry  the  concrete  during  setting.     The 


SHOP   FLOORS  321 

strength  of  these  floors  is  due  largely  to  the  concrete.  The 
thrust  may  be  calculated  by  the  formula  given  for  brick  arches. 

There  are  several  forms  of  hollow- tile  arches.  The  tile  are 
made  to  give  a  flat  arch  when  in  place.  The  fill  above  the  tile 
is  a  lean  cinder  concrete,  Fig.  345. 

Reinforced-concrete  Floors  (Fig.  347).  —  These  floors  may  be 
concrete  slabs  placed  across  steel  beams  as  shown  in  the  figure, 
or  across  reinforced-concrete  beams,  in  which  case  the  slab  forms 
the  flange  of  the  T  beam,  the  rectangular  beam  then  becoming 
its  stem.  The  design  of  these  reinforced-concrete  slabs  and 
T  beams  has  been  fully  considered  in  Chapter  XIV. 


/-Nailing  Stripe 


FIG.  347. 

The  reinforcement  may  be  either  plain  steel  rods,  twisted  or 
deformed  rods  or  any  of  the  woven  wires  or  expanded  metals 
intended  for  such  reinforcement.  When  rods  are  used  some 
reinforcing  steel  should  be  placed  at  right  angles  to  the  main 
reinforcing  rods.  These  slabs  may  have  any  of  the  previously 
described  wearing  surfaces. 

Concrete  Wearing  Surfaces.  —  This  surface  should  be  made  to 
be  as  nearly  waterproof  as  possible.  The  stone  should  not  ex- 
ceed |  in. ;  it  should  be  hard,  either  trap,  granite,  hard  blue  lime- 
stone or  similar  rock;  no  soft  rocks  are  suitable.  This  concrete 
is  frequently  i  part  cement,  i  part  sand,  and  i  part  stone;  this 
should  be  laid  upon  the  slab  before  the  cement  of  the  slab  has 
fully  set.  Where  the  slab  has  fully  set  its  face  should  be 
thoroughly  cleaned  and  then  covered  with  a  neat  cement  slush 
upon  which  the  wearing  surface  is  then  laid. 

Floors  may  be  made  practically  waterproof  by  the  thorough 
trowelling  of  the  surface.  It  may  also  be  waterproofed  by  coating 
it  with  pure  linseed  oil  thinned  with  turpentine  or  naphtha. 


322  GRAPHICS  AND  STRUCTURAL  DESIGN 

Some  concrete  floors  are  painted  but  this,  being  merely  a  surface 
effect,  soon  wears  off. 

Iron  Floors.  —  Foundry  floors,  steel-mill  charging  floors  and 
rolling-mill  floors  are  frequently  either  steel  or  iron.  These  may 
be  cast-iron  plates,  steel  plates  and  in  the  case  of  steel-mill 
charging  floors  rolled-steel  channels  placed  side  by  side  upon 
their  flange  edges. 

Where  a  smooth  surface  is  not  one  of  the  requirements  of  the 
floor,  corrugated  plates  give  a  better  footing.  Several  types  of 
corrugated  plates  and  of  plates  with  holes  in  them  are  made  by 
the  various  manufacturers  of  structural  steel. 

Steps.  —  Concrete  steps,  when  made  of  hard  stone,  wear  well 
and  give  satisfaction.  Corrugated-iron  steps  may  be  used  and 
furnished  with  one  of  the  patented  edges  to  prevent  slipping. 


CHAPTER  XX 
WALLS    AND    ROOFS 

THE  ordinary  forms  of  walls  include:  (i)  Wooden  walls  or 
sides.  (2)  Corrugated-steel  sides.  (3)  Brick,  or  brick  with  a 
concrete  face.  (4)  Stone.  (5)  Solid-concrete  walls  in  steel 
framing.  (6)  Hollow-concrete  blocks.  (7)  Reinforced  concrete 
with  expanded  metal  or  other  reinforcement. 

1.  Wooden  walls  or  sides  are  serviceable  where  the  nature  of 
the  work  carried  on  in  the  building  requires  considerable  addi- 
tional ventilation  during  part  of  the  year,  as  the  wooden  sides 
may  be  readily  removed.     The  wooden  sides  may  be  attached 
to  either  timber,  metal  or  stone  construction. 

2.  Corrugated-steel    Sides.  —  These    are   more    substantial 
than  the  wooden  ones.     The  common  sizes  of  corrugated-steel 
sheets  are  given  under  roof  coverings  and  the  comments  made 
there  apply  in  a  general  way  to  the  sides.     The  metal  for  the 
sides  is  ordinarily  a  little  lighter  than  that  on  the  roof,  being 
commonly  No.  20  or  No.  22  gauge. 

The  sheets  may  be  secured  to  girts  by  clips  or  metal  straps  or 
may  be  nailed  to  studding.  Where  the  building  must  be  kept 
warm  a  lining  of  still  lighter  corrugated  steel  may  be  placed 
inside  the  first  sheets,  an  air  space  being  left  between  the  two. 

3.  Brick.  —  The  required  thickness  of  brick  walls  for  any 
community  is  given  by  its  "  Building  Code."    An  average  state- 
ment of  such  requirements  for  a  ten-story  building  would  give 
about  the  following  wall  thicknesses. 

These  sizes  are  approximate,  being  intended  to  be  multiples 
of  brick  dimensions.  They  ordinarily  apply  to  buildings  not 
exceeding  120  ft.  in  length  and  25  ft.  in  width.  Where  the  walls 
exceed  120  ft.  between  lateral  supports  or  the  distance  between 

323 


3  24 


GRAPHICS  AND   STRUCTURAL  DESIGN 


side  walls  is  more  than  25  ft.  the  thicknesses  are  commonly 
increased  4  ins.  or  each  additional  120  ft.  or  part  thereof  in  the 
length  or  each  additional  25  ft.  or  part  thereof  in  the  width  of  the 
building. 


Numbers  of  stories. 

Estimated  height 
in  feet  to  roof. 

Thickness  of  wall 
in  inches. 

IO 

H 

12-16 

9  and  8 

40 

1  6-20 

8,  7  and  6 

80 

2O-24 

5,  4  and  3 

1  2O 

24-28 

2  and  i 

ISO 

28-32 

Brick  foundation  walls  or  footings  usually  exceed  the  walls 
immediately  upon  them  by  at  least  4  ins.  The  pressure  per 
square  inch  upon  the  brick  should  not  exceed  125  to  150  Ibs. 

Where  heavy  concentrated  loads,  such  as  trusses  or  crane  loads, 
are  transferred  through  walls  it  is  frequently  more  economical 
to  buttress  the  wall  at  these  points.  In  this  case  the  design 
should  be  treated  similarly  to  a  pillar  and  the  resultant  pressure 
should  fall  within  the  middle  third  of  the  buttress.  The  walls 
between  the  buttresses  can  then  be  made  lighter. 

Concentrated  loads  may  also  be  carried  by  steel  columns, 
either  built  in  the  wall  or  placed  clear  of  the  wall  and  inside  the 
building. 

Where  the  walls  are  supported  by  columns,  they  need  not 
exceed  12  ins.  in  thickness,  while  if  the  columns  are  not  in  the 
wall  it  should  be  made  thicker. 

Where  walls  are  cut  into  considerably  for  doors  and  windows 
their  thicknesses  should  be  increased. 

4.  Stone  walls  are  usually  made  thicker  than  brick  walls  would 
be  made  for  the  same  situation.     Few  factory  buildings  are  now 
made  of  stone  unless  very  favorably  located  to  suitable  quarries. 

5.  Solid-concrete  Walls.  —  These  are  frequently  light  walls 
placed  between  the  columns  and  steel  framing  of  a  building. 
The  objections  to  a  solid-concrete  wall  are  the  probability  of 


WALLS   AND   ROOFS 


325 


moisture  passing  through  the  concrete  or  the  wall  sweating  on 
the  inside,  also  the  liability  of  such  walls  to  crack  when  not 
reinforced  to  prevent  it.  Concrete  walls  of  this  type  are  usually 
made  thinner  than  brick  walls  would  be  made  for  the  same 
location. 

6.  Hollow-concrete  Blocks.  —  These  are  made  in  numerous 
forms.     Being  hollow  they  are  poorer  conductors  of  heat  than 
solid  walls  and  are  therefore  not  liable  to  have  moisture  con- 
densed on  the  inside.     Where  desired,  plaster   can  be  placed 
directly  on  the  blocks  without  furring  and  lathing. 

7.  Reinforced-concrete  Walls.  —  These    may   be   made    by 
running  across  the  girts  small  f -in.  channels,  having  flanges  about 


FIG.  348. 

f  in.  wide,  and  spaced  from  12  to  16  ins.  center  to  center. 
Metal  lathing  is  fastened  to  these  channels,  and  upon  this  is 
placed  a  coating  of  concrete  on  both  sides  until  the  wall  is  made 
about  2 J  ins.  thick;  see  Fig.  348. 

When,  on  account  of  heating  or  dampness,  a  hollow  wall  is 
desired,  an  inside  wall  similar  to  the  outside  one  but  plastered 


FIG.  349. 

only  on  one  side  may  be  added,  as  in  Fig.  349;  an  air  space  is 
left  between  the  two  wall  sections. 

Through  the  courtesy  of  the  American  Bridge  Company  the 
following  illustrations,  Figs.  350  to  352,  show  the  corrugated-steel 
details  recommended  by  them,  and  Figs.  353  to  359  give  their 
details  of  various  types  of  window  frames  and  sashes. 


(  Roof  Steel 


GABLE  FINISH 

WITH   PARAPET 

WALL 


Flashing  turned  int 
joints  of  brick  and 
stepped  about  everj 
2'G* 


GABLE  FINISH  FOR  STEEL  END 


Roof  steel  turned  up 
behind  vent  end  steel 

ft — .3-       FINISH  OF  VENT  ENDS 

|         w|f  Finkh  angle 


SIDE  LAP  FOR  ROOF  COR. ROOF 


GABLE  FINISH  WITH  BRICK  WALL 


TABLE  OF  CLINCH   RIVETS 


Clinch  rivets  spaced  every  C 
Rivet  always  to  go  through 
top  of  corrugation 


VALLEY    GUTTER   ;>* \ 

20  Gal.  steel  unless 
otherwise  specified 


x  18  Straps  spaced 
every  I'O  "riveted  to 
corrugated  steel 


Straps  every  4  0 

<! 

False  bottom  - 


BOX 

CORNICE 
GUTTER 


FIG.  350. 

Corrugated  steel  for  roofing  is  rolled  from  a  sheet  30  inches  wide  in  the  flat,  27  £  inches 
when  rolled  one  edge  up  and  one  edge  down.  Laid  with  i£  inches  of  corrugated  lap  will 
cover  24  inches  of  roof. 

When  ordering  state  distinctly  that  the  sheeting  is  for  roofing,  that  it  is  to  be  27  £  inches 
wide  after  corrugating,  and  that  the  corrugations  are  to  be  |  inch  deep.  State  whether  the 
sheeting  is  to  be  plain  or  corrugated,  and  also  if  it  is  to  be  painted.  Specify  the  gauge. 

Wherever  possible  order  sheets  in  even  feet  lengths  and  to  span  two  purlin  spaces. 
Allow  6  inches  for  end  laps  in  roofs  of  6  inches  pitch  and  8  inches  for  end  laps  in  roofs  of 
4  inches  pitch.  In  roofs  of  less  than  4  inches  pitch  allow  8  inches  for  end  laps  and  lay  with 
Slater's  cement. 

SIZES  OF  GUTTERS  AND  CONDUCTORS. 


Span  of  roof,  feet. 

Gutters, 
inches. 

Conductor,  inches. 

Up  to  50  
50  to  70  
70  to  100  

6 
7 
8 

4  every  40  feet 
5  every  40  feet 
5  every  40  feet 

These  are  made  of  No.  24  galvanized  steel  unless  otherwise  specified, 
should  slope  at  least  i  inch  in  15  feet. 


Hanging  gutters 


WALLS   AND   ROOFS 


327 


PLAIN  FINISH 


FLASHED  FINISH 


orv  !— 6- 

,|    1     Clo.fcgJU, 

j    fr    INSJDB  CORNER 


CAPPING 


OUTSIDE  CORNER 
CAPPING 


ClipeuidBalu 


FIG.  351- 

Corrugated  steel  for  siding  is  rolled  from  steel  sheets  28  inches  wide  in  the  flat,  and  is 
26  inches  wide  when  rolled  with  both  edges  down.  When  laid  with  a  side  lap  of  one  corru- 
gation it  will  cover  24  inches  of  side.  When  ordering  state  distinctly  that  the  sheeting  is 
for  siding,  that  it  is  to  be  26  inches  wide  after  corrugating,  and  that  the  corrugations  are 
to  be  |  inch  deep.  State  whether  the  sheeting  is  to  be  galvanized  or  black,  and  if  it  is  to 
be  painted;  also  give  gauge.  Sheets  should  be  ordered  in  even  feet  lengths  and  to  span  two 
purlin  spaces,  wherever  possible.  Allow  4  inches  for  end  laps.  If  side  laps  are  to  be 
riveted  space  closing  rivets  about  12  inches  apart.  Provide  roller  guides  and  door  stops 
to  hold  doors  securely  in  place  when  open  or  shut. 


GRAPHICS  AND   STRUCTURAL  DESIGN 


SECTION  THROUGH      SECTION  THROUGH 
FLAT  OR  CRIMPED  CORRUGATED 

STEEL  SHUTTER  STEEL  SHUTTER 


Roof  {Steel 


Use  c.ngle  uprights  at         ^^  " 
splice  joints  at  louvres 

.^-Hfbracketsat 

splice  joints  p- 

Gauge  of  metal 
22  unless  specified 


1 V*  i/Strap  at 


BERLIN 
LOUVRES 


FIG.  352. 


Shutters  are  made  from  6  feet  to  10  feet  long,  all  standard  width.  Two  hinges  are  used 
on  shutters  up  to  8  feet  long,  while  larger  sizes  are  given  three  hinges. 

Louvres  of  the  Berlin  type  are  made  of  No.  24  gauge  steel.  The  maximum  length  is 
4  feet  i£  inches.  The  end  lap  is  from  \  inch  to  \  inch. 

Steel  for  these  louvres  should  be  ordered  n  inches  wide.  The  uprights  have  %6-inch 
diameter  holes  for  J-inch  oval  screw-head  bolts,  \  inch  long. 

Schifner  Louvres.  The  maximum  lengths  of  these  are  7  feet  o  inches,  without  lap. 
The  steel  should  be  ordered  n  inches  wide.  The  holes  are  %e  inch  in  diameter  for  1-inch 
diameter  bolts  with  oval  screw  heads  and  i  inch  long. 


WALLS   AND   ROOFS 


329 


.  _ .    _  -W,=Gla8s,  Muntins  and  4>$"  -  - 
PLAN 

FIG.  353. 

The  design  shown  in  Fig.  353  is  for  fixed  sash  for  monitors;  for  swing  monitor  sash  cut 
stops  off  as  shown  by  dotted  lines,  and  omit  head  stop  on  the  inside.  Make  frames  and 
sash  of  white  pine,  excepting  spiking  and  blocking  pieces,  which  should  be  spruce,  hemlock 
or  Norway  pine,  planed  on  all  exposed  sides.  For  swing  sash  order  two  trunnions  for  each 
sash,  and  call  for  lever  operating  device. 


33° 


GRAPHICS   AND   STRUCTURAL   DESIGN 


Cor.  Steel 


/Purlin          Tin  net's  Nails        N-* 

^•_— — :r-^ i  /Y\  o*  I 

"F" 


Muntfn 


Bo 


Top  Kail 


»m"Rafl 


^Cor.  Steel  l-^'Wund 


ELEVATION 


-W=Glass., 


PLAN 


FIG.  354- 


The  design  shown  in  Fig.  354  is  for  continuous  fixed  sash  in  corrugated  steel  sides.     Make 
sash  and  sill  of  white  pine,  planed  on  all  exposed  sides. 


WALLS  AND   ROOFS 


331 


COr'Steel        Las  Screws  yfkl*> 

r^!^P^t^r8- 
Tn         53-, 

~e« 


=  Gloss,  Muntins 
PLAN 


FIG.  355. 


The  design  shown  in  Fig.  355  is  for  sliding  sash  in  corrugated  steel  sides.     The  stop 
used  for  roller  track  is  to  be  hard  wood,  other  wood  the  same  as  recommended  for  Fig.  353. 


332 


GRAPHICS  AND   STRUCTURAL  DESIGN 


Glass-  --  -Glass-  -         -Glass-  *nd-* 
W  -  Glass,  Muntins, and  4}£"=4'o  Mast  **j       f 
PLAN 

FIG.  356. 


The  design  shown  in  Fig.  356  is  for  a  window  frame  with  counterbalanced  sash  in  corru- 
gated steel  sides.  Neither  dimension  of  sash  exceeds  4  feet  o  inches.  The  wood  is  recom- 
mended the  same  as  for  Fig.  353. 


WALLS   AND    ROOFS 


333 


|Mi 

rlaening 

•1--*-=,— 

Purln 
.---6 

1 

£ 

[_1 

5 

Top  Rail 
Muntin 

1 

;i||,. 

rf&t 

Meeting  Rai 

tfi 

'* 

M 
* 

Tn 

Dimem 
bynui 

on  of  sash  dt 
her  and  eize 

Muniin 

ormined 
>f  lighW 

•/. 

f-i 

^Purlin 

P 

Bottom  Rail 

—  V 

r 

iJUllllll  U  1 

--r*t  — 

Purlin 

..  _._^ 

imtmil 

.  ./  . 

^__—  --Cor 

Steel 

window  in  field 


ELEVATION 


Drip/111  \,*««8ill 
1*-X  roiuid  |T  H  x^X  1-W  Screws 

SECTION 


y^2C^^BM^S£«Bj»C 

f~—  W  ~Gla86,MuDtins,and  4&Wl  Min.  - 
PLAN 

FIG.  357. 

The  design  shown  in  Fig.  357  is  for  a  window  frame  with  counterbalanced  sash  in  corru- 
gated steel  sides.  In  this  design  one  dimension  is  not  less  than  4  feet  i  inch.  The  wood 
used  is  similar  to  that  specified  for  Fig.  353. 


«K'  '  . 

-Glass^VH^2 


334 


GRAPHICS  AND   STRUCTURAL  DESIGN 


&xl&  Lag  Screw 
2"x3"Block 


Dimensi  n  of  sash  d« 
by  number  and  size 


Mantins,  and4K*=i'0*Max. 
PLAN 

FIG.  358. 


WALLS   AND    ROOFS 


335 


i-  Flashing 
Purlin 

In'. 

"I*1  T'T 

13£*  Lag  Sore  w 
2x3"Block 

s?  r 

-----  4— 

r—  --^p—  —r—---^ 

FT-                                                -ffy            =' 

sssaa^                   ^  yj 

-- 

-1- 

T|± 

ii._t 

- 

i 

-^  Mortised 

Top  Rail    I 

1 

Pulley 

« 

-  1 

i               I 

1 

-, 

Blunt  in 

i    4" 

1 

j 

M 

q 

*     i 

11 

? 

1 

: 

•;         •? 

.9 

Mectii 

ijjRail 

±3     -:  '^r 

i 

f! 

x      ! 

r 

Dimensi 

n  of  sauh  det 

rmined 

i 

* 

.3 

by  uun 

"  -r  and  size  < 

f  lighta 

3 

1 

0 

5 

- 

l 

vsi-« 

1 

1 

7. 

*T 

i 

i 

1 

1 

1 

1 
1 

•III 

o 

1 

1 

i 

1 

t 

1 

1 

Dot  to 

in  Rail 

-i'4 

•vl^f-^ 

1 

itapai 

=»-               =»                     ""-  'fijtoptff    •-       *- 
Purlin                 "                          \                  D^VVPrX?1*"1**8"1 

„_ 

Cor-Steel                                                                       ^roUnd    #     **8S1U 

~     >»» 

ELEVATION                                                                SECTION 

X  * 


I    h-5;^2X'  jy  JiX*  2; 

IJi"  '          J*—  W»Gla88,  Muntina,  and  l#*l'l"Min.    -*{ 
PLAN 

FIG.  359- 

The  designs  shown  in  Figs.  358  and  359  are  for  window  frames  with  double  hung  weighted 
sash  in  corrugated  steel  siding.  In  Fig.  358  neither  dimension  of  the  sash  exceeds  4  feet 
p  inches,  while  in  Fig.  359  one  dimension  is  not  less  than  4  feet  i  inch.  The  sills  and  cas- 
ings should  be  made  of  white  pine;  the  jambs  and  parting  strips  should  be  made  of  hard 
pine;  while  spiking  pieces  and  blocks  should  be  made  of  hemlock  or  Norway  pine,  planed 
on  all  exposed  sides. 


GRAPHICS  AND   STRUCTURAL   DESIGN 
CRANE  CLEARANCES  AND  WEIGHTS 

I  .  Not  less  than  3 " 


-Not  lose  thin  2" 

Ar—       3 


Truck  Wheel-bsatf 


Capac- 
ity, tons 
(2000 
Ibs.). 

Span. 

A. 

B. 

Wheel 
base. 

Weight  of  bridge, 
2  girders. 

Weight  of 
trolley. 

Rail. 

Ft. 

Inches. 

Ft.    Ins. 

Ft.  Ins. 

Pounds. 

Pounds. 

Lbs.  per 
yard. 

( 

40 

8 

4     9 

8     0 

8,600-   13,000 

4,500 

) 

5) 

60 

8 

4     9 

8     o 

16,500-   2I,OOO 

to 

[    40 

80 

8 

4     9 

8     o 

28,OOO-  33,000 

5,000 

) 

( 

40 

8 

5     o 

8     6 

12,300-   l6,OOO 

6,000 

) 

IO  < 

60 

8 

5     o 

8     6 

2O,OOO-   24,000 

to 

40 

( 

80 

8 

5     o 

9     o 

32,000-  37,000 

8,000 

) 

( 

40 

8 

5     6 

9     o 

14,000-  23,000 

9,000 

; 

15  ) 

60 

8 

5     6 

9     6 

23,000-  33,000 

to 

50 

80 

9 

5     6 

IO      O 

37,000-  48,000 

10,000 

) 

( 

4-O 

6     o 

23,000—  28,000 

10,000 

) 

20  5 

if-W 

60 

_ 

6     o 

30,000—  37,000 

to 

So 

I 

80 

9 

6     o 

40,000-  52,000 

12,000 

$ 

( 

40 

9 

6     6 

IO      O 

19,000-  34,000 

12,000 

) 

2S] 

60 

9 

6     6 

10     6 

29,000-  45,000 

to 

f    60 

i 

80 

9 

6     6 

II       0 

45,000-  61,000 

15,000 

) 

( 

40 

9 

7     o 

II       O 

34,000-  37,000 

14,000 

} 

SO] 

60 

9 

7     o 

ii     6 

44,000-  49,000 

to 

[    60 

( 

80 

9 

7     o 

12       O 

58,000-  66,000 

17,000 

) 

( 

40 

9 

8     o 

12       6 

43,000-  49,000 

16,000 

^ 

40  < 

60 

9 

8     o 

13     o 

60,000-  63,000 

to 

[    80 

1 

80 

9 

8     o 

13     3 

70,000-  82,000 

20,000 

) 

( 

40 

9 

8     6 

12       6 

48,000-  57,000 

24,000 

) 

50 

60 

9 

8    6 

13     o 

66,000-  73,000 

to 

>  IOO 

( 

80 

9 

8     6 

13     6 

85,000-  95,000 

30,000 

) 

( 

40 

10 

9     o 

15       0 

78,000 

60] 

60 

IO 

9     o 

15     3 

70,000-  95,000 

32,000 

f 

80 

10 

9     o 

IS     6 

100,000-126,000 

IOO 

( 

40 

10 

9     6 

15     6 

101,000 

75 

60 

IO 

9     6 

IS     6 

80,000-120,000 

40,000 

to 

/ 

80 

IO 

9     6 

IS     6 

120,000-144,000 

( 

40 

12 

IO      O 

IS     6 

134,000 

100  < 

60 

12 

10      0 

IS     6 

94,000-161,000 

56,000 

15° 

I 

80 

12 

10      0 

IS     6 

125,000-187,000 

WALLS   AND    ROOFS  337 

The  maximum  wheel  loads  are  approximately  25  per  cent  of 
the  bridge  weight  plus  50  per  cent  of  the  combined  weights  of  the 
trolley  and  crane  capacity. 

The  dimensions  of  windows  depend  upon  the  light  required 
and  the  size  of  the  panes  used.  In  shop  construction  the  greatest 
possible  amount  of  light  is  wanted  but  without  sun.  On  this 
account  a  northern  light  is  preferred.  In  one-story  buildings  this 
leads  to  saw-tooth  or  similar  construction. 

The  usual  sizes  of  common  glass  can  be  obtained  in  any  hand- 
book on  Building  Construction.  The  regular  sizes  of  glass  range 
from  6X8  ins.  to  44  X  72  ins.;  the  dimensions  up  to  16  ins. 
varying  by  inches,  while  those  above  16  ins.  vary  by  2  ins.  and 
are  the  even  inches  only. 

Single-thick  glass  is  about  TXg  in.  thick. 

Double-thick  glass  is  about  ^  in.  thick. 

Wired  glass  has  the  advantage  of  not  falling  out  when  cracked, 
thus  affording  increased  fire  protection.  Translucent  fabric  is 
a  substitute  for  glass  that  has  some  advantages.  It  cuts  out  a 
little  light  but  when  a  sufficient  amount  is  used  it  affords  perfect 
lighting.  It  is  impervious,  elastic  and  burns  with  difficulty. 

The  subject  of  methods  of  lighting  is  an  interesting  one  and 
should  be  given  careful  consideration  in  buildings  of  any  impor- 
tance. According  to  Ketchum  it  is  common  to  specify  that  10 
per  cent  of  the  exterior  surface  of  ordinary  mill  buildings  and  25 
per  cent  of  machine  shops  and  similar  buildings  should  be  glazed, 
this  as  a  rule  being  divided  between  windows  and  skylights. 

ROOF  COVERINGS 

Corrugated-steel  Roofing.  —  This  is  one  of  the  commonest 
kinds  of  roofing  materials.  It  is  made  by  rolling  the  plain  sheets 
with  corrugations. 

These  sheets  range  from  No.  16  to  No.  28,  U.  S.  Standard 
gauge  (0.063  m-  to  0.016  in.).  The  corrugations  range  from 
T3g  in.  to  2.5  ins.  A  few  companies  roll  sheets  with  corrugations 
of  5,  3  and  2  ins. 


338 


GRAPHICS  AND  STRUCTURAL  DESIGN 


The  sizes  most  frequently  used  are  gauge  Nos.  20  and  24,  with 
2|-in.  corrugations,  f  in.  deep.  The  lengths  of  sheets  range  from 
6  ft.  to  12  ft.  A  common  length  is  8  ft. 

Roofs  covered  with  corrugated  steel  should  not  have  a  slope 
of  less  than  3  ins.  in  12  ins.  According  to  Kidder's  Handbook 
they  should  be  supported  on  about  the  following  spans: 


Gauge. 

Span  in  feet. 

Gauge. 

Span  in  feet. 

24 
22  and  20 

2-2.5 
2-3 

18 
16 

4-5 
5-6 

The  lap  at  the  end  of  the  sheet  should  range  from  3  to  6  ins. 
according  to  the  slope. 

Slope  i  to  3 Lap  3  ins. 

Slope  i  to  4 Lap  4  ins. 

Slope  i  to  8 Lap  5  ins. 

The  side  laps  vary  from  one  to  two  corrugations;  a  lap  of  i| 
corrugations  makes  a  very  good  joint;  see  Fig.  360. 


FIG.  360. 

Corrugated  roofing  is  secured  to  the  building  at  the  purlins 
by  nails  when  nailing  strips  are  used,  and  by  hoop-iron  clips  or 
bands  when  secured  directly  to  the  purlins;  see  Figs.  350  to  352. 
The  riveting  and  nailing  should  be  done  at  the  top  of  the  corru- 
gations. 

The  Pencoyd  Handbook  gives  the  following  crippling  load  in 
pounds  per  square  foot: 

ooo  X  /  X  b  X  d\ 


W 


/Q8, 


t  =  thickness  of  the  metal  in  inches. 

b  =  center  to  center  of  corrugations  in  inches. 

d  =  depth  of  corrugations  in  inches. 


WALLS  AND   ROOFS 


339 


length  of  span  in  inches.  The  same  handbook  gives 
the  following  safe  loads  in  pounds  per  square  foot  on 
corrugated  steel  of  2^-in.  corrugations  and  f  in.  deep: 


Span  in  feet. 

Gauge 

No. 

3 

4 

5 

6 

2O 

59 

44 

36 

30 

22 

48 

36 

2Q 

24 

24 

37 

28 

22 

19 

26 

3i 

23 

18 

16 

Purlins  should  be  spaced  for  a  roof  load  of  not  less  than  30  Ibs. 
per  sq.  ft.;  spacing  for  lighter  loads  results  in  injury  to  the  cor- 
rugated-steel joints  when  the  roof  is  walked  upon.  The  sizes 
most  frequently  used  are  No.  20  for  the  roof  and  No.  22  for 
the  sides.  The  corrugated  steel  should  be  ordered  in  lengths 
sufficient  to  cover  two  purlins.  The  laps  should  be  painted 
before  being  riveted.  Galvanized  steel  will  take  paint  better 
after  weathering  a  while. 

When  the  buildings  are  lined,  ij  by  f-in.  corrugated  steel  is 
frequently  used.  When  lined  the  corrugated  steel  is  better 
nailed  to  nailing  strips. 

The  greatest  objection  to  the  corrugated-steel  roofing  is  the 
fact  that  it  sweats.  In  heated  buildings  or  in  rooms  with  moist 
air  the  moisture  condenses  on  the  under  side  of  the  roof  and 
drops  to  the  floor,  possibly  striking  machinery  or  materials  it 
may  injure.  A  common  method  of  overcoming  this  condensation 
is  to  cover  the  purlins  with  galvanized  poultry  netting,  running 
the  netting  from  the  eaves  on  one  side,  over  the  purlins  and  down 
to  the  eaves  on  the  other  side.  The  netting  is  secured  to  the 
eaves  and  the  several  widths  of  netting  are  woven  together  along 
the  edges.  Upon  this  netting  are  then  laid  one  or  two  layers  of 
asbestos  paper,  and  on  this  one  or  two  layers  of  tar  or  other 
paper  impervious  to  moisture;  finally  the  corrugated  steel  is 
placed  on  these.  The  netting  is  60  ins.  wide  and  weighs  10  Ibs. 


340 


GRAPHICS  AND   STRUCTURAL  DESIGN 


per  square  of  100  sq.  ft.  The  asbestos  paper  is  about  T^  in. 
thick.  The  asbestos  paper  prevents  the  tar  or  other  paper  from 
taking  fire,  thus  making  an  excellent  fireproof  roof. 


Weight  of  corrugated  steel, 
in  pounds  per  100  sq.  ft. 

Gauge. 

Black. 

Galvanized. 

20 

165 

182 

22 

138 

154 

24 

III 

127 

26 

84 

99 

FIG.  361. 


Slate  Roofing.  —  Slate  makes  a  very  satisfactory  roof  cover- 
ing. The  best  slates  have  a  somewhat  metallic  appearance,  do 

not  absorb  water  and  are  strong. 
Generally  the  strongest  slates  are 
the  best.  The  sizes  range  from 
6  X  12  ins.  to  16  X  26  ins.  and  in 
thickness  from  \  in.  to  J  in.  The 
most  frequently  used  sizes  run  from 
6  X  12  ins.  to  12  X  18  ins.,  and 
are  generally  T3^  in.  thick.  In  slate 
roofs  the  slope  should  not  be  less  than  i  in  4.  The  usual  lap  is 
a  double  lap  of  3  ins. ;  that  is,  the  upper  3  ins.  of  the  first  slate 
are  covered  by  the  third  slate;  see  Fig.  361. 

As  slate  breaks  easily  under  shock,  the  roof  must  be  designed 
to  deflect  but  slightly  under  loading. 

The  slates  are  commonly  laid  on  sheathing;  this  maybe  either 
plain  or  tongued-and-grooved  boards,  their  thickness  varying 
with  the  purlin  spacing  and  the  load.  The  sheathing  is  covered 
with  tar  paper  or  with  waterproof  paper  or  felt.  The  slates  may 
be  laid  on  roofing  laths;  these  are  2  to  3  ins.  wide  and  i  in.  to 
ij  ins.  thick;  they  are  spaced  on  the  rafters  to  suit  the  nails  in 
the  slates.  At  gutters,  valleys  and  ridges  the  slates  should  be 
laid  in  cement  and  are  sometimes  entirely  laid  that  way. 


WALLS    AND   ROOFS  341 

The  following  are  examples  of  slate  roofing: 

Trusses  14  ft.  6  ins.  center  to  center.  No  purlins.  Sheathing 
3-in.  yellow-pine  plank.  The  planks  were  29  ft.  long  and  joints 
were  broken  on  alternate  trusses. 

Another  roof  whose  purlins  were  5  ft.  8  ins.  center  to  center 
had  2-in.  yellow-pine  sheathing. 

Although  slate  is  an  expensive  roofing  material,  it  can  reason- 
ably be  expected,  with  ordinary  repairs,  to  last  from  25  to  30 
years.  While  not  resisting  great  heat  (it  cracks  and  disinte- 
grates, and  then  exposes  the  sheathing  to  the  fire),  slate  itself 
does  not  ignite,  and  thus  makes  a  fairly  good  fireproof  roof. 

Clay  tiles  make  a  thoroughly  fireproof  roof.  They  are,  however, 
very  heavy  and  expensive.  Several  makes  of  tiles  are  also  molded 
in  glass,  thus  permitting  skylights  to  be  very  readily  placed. 
Their  artistic  effect  is  good. 

Concrete  Roofs.  —  Concrete,  reinforced  by  expanded  metal 
or  steel  rods,  is  also  much  used  for  roofs.  The  purlins  may  be 
spaced  from  5  to  7  ft.,  or  they  may  be  omitted  altogether,  the 
concrete  slabs  being  placed  directly  on  the  trusses  or  on  concrete 
beams  or  rafters.  When  the  slope  is  steep,  slate  may  be  nailed 
directly  to  the  concrete;  this  should  be  done  within  a  couple  of 
weeks  after  placing  the  concrete;  or  if  the  slope  is  slight  a  stand- 
ard slag  roof  may  be  used. 

The  following  is  an  example  of  a  reinforced-concrete  roof :  The 
roof  slab  was  4^  ins.  thick;  it  was  reinforced  with  ^-in.  square 
bars,  spaced  5  ins.  center  to  center;  there  were  no  longitudinal 
beams;  the  transverse  girders  were  14  ft.  center  to  center. 
Wooden  nailing  strips  were  embedded  in  the  concrete  slabs; 
these  secured  a  five-ply  waterproofing  course  upon  which  a  stand- 
ard slag  roof  was  placed. 

Slag  or  Gravel  Roofing.  —  This  roof  covering  may  be  placed 
on  either  wooden  sheathing  or  on  a  reinforced-concrete  roof. 
Such  a  roof  when  on  wooden  sheathing  is  commonly  made  by 
covering  the  wood  with  dry  felt  paper  and  on  this  placing  from 
three  to  five  layers  of  tarred  or  asphaltic  felt.  The  layers  of 


342 


GRAPHICS   AND   STRUCTURAL   DESIGN 


paper  are  lapped  the  same  as  slate,  exposing  from  6  to  10  ins.  of 
each  sheet. 

Another  method  is  to  lay  rosin-sized  sheathing  paper  weighing 
not  less  than  6  Ibs.  per  100  sq.  ft.;  upon  this  lay  two  ply  of 
tarred  felt  lapping  17  ins.  and  weighing  not  less  than  14  Ibs.  per 
100  sq.  ft.  single  thickness.  These  are  then  covered  with  a  spread- 
ing of  pitch,  and  finally  three  additional  layers  of  felt  are  laid  and 
similarly  coated  with  pitch.  On  this  layer  of  pitch  the  slag  or 
gravel  is  spread;  if  slag,  it  should  be  crushed  smaller  than  f  in. 
but  larger  than  J  in.  Gravel  should  be  screened  and  both  slag 
and  gravel  should  be  clean  and  dry.  When  laid  on  cement 
roofing  the  rosin  sheathing  paper  should  be  omitted  but  the 
cement  should  be  coated  with  pitch.  Although  a  slag  or  gravel 
roof  is  not  fire  proof,  tests  have  shown  its  fire-resisting  qualities 
to  be  good.  It  protects  the  wooden  sheathing  better  than  slate. 


SLOPE  OF  ROOFS 

The  slope  of  a  roof  depends  upon  that  required  by  the  roofing 
material  used.     The  following  is  taken  from  Kidder's  Handbook. 


Roofing  material. 

Slope  of  rafter, 
inches  per  foot. 

Slate                             

8 

Tiles  (interlocking) 

7 

Tin  shingles  (painted) 

6 

Cedar  shingles 

6 

Corrugated  steel 

•2 

Ready  roofing 

I 

In  the  case  of  slate  the  permissible  slope  will  vary  with,  the 
weight  and  size  of  the  slate. 

Largest  sizes  of  slate Slope  5  ins.  per  ft. 

Medium  sizes  of  slate   Slope  6  ins.  per  ft. 

Smallest  sizes  of  slate Slope  8  ins.  per  ft. 

Roofs  having  a  slope  of  J  in.  to  f  in.  per  ft.  constitute  what  are 
termed  flat  roofs,  and  are  generally  covered  with  tar  and  gravel, 
asphalt,  ready  roofing  or  tin  with  lock  and  soldered  joints. 


WALLS  AND   ROOFS  343 

Pitch  roofs  should  be  given  a  slope  not  exceeding  2  ins.  per  ft. 

Rafters  or  their  equivalents  should  be  spaced  not  over  5  ft. 
when  ordinary  sheathing  of  i-in.  boards  is  used. 

The  following  are  examples  of  roofs : 

Pennsylvania  Steel  Co.  —  Purlins  6  ft.  center  to  center. 
Sheathing  i|-in.  tongued  and  grooved  hemlock.  Purlins  4  X 
lo-in.  yellow  pine.  Slag  roofing  trusses  20  ft.  center  to  center. 

A  Bridge  Shop  Roof.  —  Composition  roofing.  Purlins  spaced 
about  7  ft.  6  ins.  center  to  center.  Sheathing  2  ins.  thick. 

Slate  Roof  on  Tar  Paper.  —  Purlins  spaced  about  4  ft.  center 
to  center.  Sheathing  i|  in. 

Slate  roof  on  2-in.  sheathing.  Purlins  spaced  about  8  ft.  center 
to  center. 

Slate  roof  on  2|-in.  plank.  Purlins  about  8  ft.  6  ins.  center  to 
center. 

Five-ply  Slag  Roofing.  —  Sheathing  i|-in.  spruce  on  purlins 
spaced  about  5  ft.  6  ins.  center  to  center. 

Asphalt  Roofing.  —  Purlins  spaced  5  ft.  6  ins.  center  to  center. 
Sheathing  if-in.  tongued  and  grooved  boards. 


CHAPTER   XXI 


SPECIFICATIONS    FOR   STRUCTURAL   STEEL 

WORK 

MATERIALS 

1.  Process  of  Manufacture.  —  Steel  may  be  made  by  either  the  open- 
hearth  or  the  Bessemer  process. 

NOTE.  —  For  the  more  important  work  rolled  steel  is  preferably  made  by  the 
basic  open-hearth  process,  as  this  process  permits  the  elimination  of  the  greater 
portion  of  the  phosphorus  and  a  small  percentage  of  the  sulphur  contained  in  the 
pig  iron  and  scrap  from  which  the  steel  is  made.  Steel  for  castings  is  commonly 
made  by  the  acid  open-hearth  process. 

2.  Chemical  and  Physical  Requirements.  — 


Chemical  and  physical  requirements. 

Structural 
steel. 

Rivet  steel. 

Steel  castings. 

Phosphorus,  maximum  
Sulphur,  maximum  
Ultimate    tensile    strength    in 
Ibs.  per  sq.  in.     See  No.  3  .... 
Elongation,  min.  per  cent  in  8 

0.04% 
0-05% 
Desired 
60,000 
1,500,000 

0.04% 
0.04% 
Desired 
50,000 
1,500,000 

0.05% 
0-05% 

Not  less  than 
65,000 

ins.,  see  Fig.  i.     See  No.  4.... 
Elongation,  min.  per  cent  in  2 
ins.,  see  Fig.  2 

Ult.  ten.  str. 

f>O7 

Ult.  ten.  str. 

_QO/ 

Character  of  fracture  

22  /Q 
Cftlr-r, 

QiTb-xr 

Jo/o 

Cold  bends  without  fracture  .... 

(  180°  flat 
\  See  No.  5 

180°  flat  ) 
See  No.  6  f 

granular 
90°,  d  =  3  t 

NOTE.  —  The  effect  of  phosphorus  is  to  make  the  steel  cold  short,  brittle  when 
cold;  while  the  sulphur  makes  it  hot  short,  brittle  when  hot. 

3.  Allowable   Variations.  —  Tensile   tests   of   steel   will   be   considered 
satisfactory  if  showing  an  ultimate  strength  within  5000  pounds  of  that 
desired. 

4.  Modifications  in  Elongation.  —  For  material  less  than  T5^  inch  thick 
a  deduction  of  2^  per  cent  will  be  allowed  in  the  elongation  for  each  Tag  inch 
the  material  is  under  T5^  inch.     For  material  more  than  f  inch  thick  a  de- 

344 


MATERIALS  345 

duction  of  i  per  cent  may  be  allowed  for  each  |  inch  the  material  exceeds 
f  inch.  For  pins  and  rollers  exceeding  3  inches  in  diameter  the  elongation 
in  8  inches  may  be  5  per  cent  below  that  given  in  No.  2. 

5.  Bending  Tests.  —  These  tests  may  be  made  either  by  pressure  or  by 
blows.     Plates,  shapes  and  bars  less  than  i  inch  thick  shall  bend  as  specified 
in  No.  2. 

Full-sized  material  for  eyebars  and  other  steel  i  inch  thick  and  over, 
tested  as  rolled,  shall  bend  cold  180  degrees  around  a  pin  whose  diameter  is 
twice  the  thickness  of  the  bar.  It  must  show  no  fracture  on  the  outside 
of  the  bend. 

Angles  I  inch  and  less  in  thickness  shall  open  flat,  angles  \  inch  and  less 
in  thickness  shall  bend  shut,  cold,  under  blows  of  a  hammer,  without  signs 
of  fracture.  This  test  will  be  made  only  when  required  by  the  inspector. 

6.  Nicked  Bends.  —  Rivet  steel,  when  nicked  and  bent  around  a  bar  the 
same  size  as  the  rivet  rod,  shall  give  a  gradual  break  and  a  fine,  silky, 
uniform  fracture. 

7.  In  order  that  the  ultimate  strength  of  full-sized  annealed  eyebars  may 
meet  the  requirements  of  paragraph  65  the  ultimate  strength  in  test  speci- 
mens may  be  determined  by  the  manufacturers;    all  other  tests  than  those 
for  ultimate  strength  shall  conform  to  the  requirements  in  paragraph  2. 

8.  The  yield  point,  as  indicated  by  the  drop  of  the  beam,  shall  be  recorded 
in  the  test  reports. 

9.  Chemical  Analyses.  —  Chemical  determinations  of  the  percentages 
of  carbon,  phosphorus,  sulphur  and  manganese  shall  be  made  by  the  manu- 
facturer from  a  test  ingot  taken  at  the  time  of  the  pouring  of  each  melt  of 
steel  and  a  correct  copy  of  such  analysis  shall  be  furnished  to  the  engineer 
or  his  inspector.     Check  analyses  shall  be  made  from  the  finished  material, 
if  called  for  by  the  purchaser,  in  which  case  an  excess  of  25  per  cent  above 
the  required  limits  will  be  allowed. 


1. 

About 
2" 


—    About  18 •* 

Parallel  section  i,,» 


not  less  than  9 " 


±t 


FIG.  i. 


10.  Form  of  Specimens.  —  (a)  Plates,  Shapes  and  Bars.  —  Specimens 
for  tensile  and  bending  tests  for  plates,  shapes  and  bars  shall  be  made  by 
cutting  coupons  from  the  finished  product,  which  shall  have  both  faces 
rolled  and  both  edges  milled  to  the  form  shown  in  Fig.  i;  or  with  both 


346  GRAPHICS  AND 'STRUCTURAL  DESIGN 

edges  parallel;  or  they  may  be  turned  to  a  diameter  of  f  inch  for  a  length 
of  at  least  9  inches,  with  enlarged  ends. 

(b)  Rivets.  —  Rivet  rods  shall  be  tested  as  rolled. 

(c)  Pins  and  Rollers.  —  Specimens  shall  be  cut  from  the  finished  rolled 
or  forged  bar  in  such  a  manner  that  the  center  of  the  specimen  shall  be 
i    inch  from   the   surface  of  the  bar.    The  specimen  for  the  tensile  test 
shall  be  turned  to  the  form  shown  by  Fig.  2.    The  specimen  for  bending 
test  shall  be  i  inch  by  f  inch  in  section. 


FIG.  2. 

11.  Steel  Castings.  —  The  number  of  tests  will  depend  on  the  number 
and  importance  of  the  castings.     Specimens  shall  be  cut  cold  from  coupons 
molded  and  cast  on  some  portion  of  one  or  more  castings  from  each  melt 
or  from  the  sink  heads,  if  the  heads  are  of  sufficient  size. 

The  coupon  or  sink  head,  so  used,  shall  be  annealed  with  the  casting  before 
it  is  cut  off.  Test  specimens  shall  be  of  the  form  prescribed  for  pins  and 
rollers. 

12.  Annealed  and  Unannealed  Specimens.  —  Material  which  is  to  be 
used  without  annealing  or  further  treatment  shall  be  tested  in  the  condition 
in  which  it  comes  from  the  rolls.     When  material  is  to  be  annealed  or 
otherwise  treated  before  use,  the  specimens  for  tensile  tests  representing 
such  material  shall  be  cut  from  properly  annealed  or  similarly  treated  short 
lengths  of  the  full  section  of  the  bar. 

13.  Number  of  Tests.  —  At  least  one  tensile  and  one  bending  test  shall 
be  made  from  each  melt  of  steel  as  rolled.     In  event  of  the  material  rolled 
from  one  melt  varying  in  thickness  by  f  inch  or  more  a  test  shall  be  made 
from  the  thickest  and  the  thinnest  material  rolled. 

14.  Finish.  —  Finished  material   shall   be  free  from   injurious   seams, 
flaws,  cracks,  defective  edges,  or  other  defects,  and  have  a  smooth,  uniform, 
workmanlike  finish.     Plates  36  inches  and  under  in  width  shall  have  rolled 
edges. 

15.  Stamping.  —  Every  finished  piece  of  steel  shall  have  the  melt  number 
and  the  name  of  the  manufacturer  stamped  or  rolled  upon  it.     Steel  for 
pins  and  rollers  shall  be  stamped  on  the  end.     Rivet  and  lattice  steel  and 


MATERIALS  347 

other  small  parts  may  be  bundled  with  the  above  marks  on  an  attached 
metal  tag. 

16.  Defective   Material.  —  Material,  which,  subsequent  to  the  above 
tests  at  the  mills,  and  its  acceptance  there,  develops  weak  spots,  brittle- 
ness,  cracks,  or  other  imperfections,  or  is  found  to  have  injurious  defects, 
will  be  rejected  at  the  shop  and  shall  be  replaced  by  the  manufacturer  at 
his  own  cost. 

17.  Allowable  Variation  in  Weight.  —  A  variation  in  cross  section  or 
weight  in  the  finished  members  of  more  than  2\  per  cent  from  that  specified 
shall  be  sufficient  cause  for  rejection. 

1 8.  Cast  Iron.  —  Except  where  chilled  iron  is  specified,  castings  shall 
be  made  of  tough  gray  iron,  with  sulphur  not  over  o.io  per  cent.     They 
shall  be  true  to  pattern,  out  of  wind  (perfectly  straight  or  flat)  and  free  from 
flaws  and  excessive  shrinkage.     If  tests  are  demanded  they  shall  be  made 
on  the  "  Arbitration  Bar  "  of  the  American  Society  of  Testing  Materials, 
which  is  a  round  bar  i  \  inches  in  diameter  and  1 5  inches  long.     The  trans- 
verse test  shall  be  on  a  supported  length  of  12  inches  with  the  load  at  the 
middle.     The  minimum  breaking  load  so  applied  shall  be  at  least  2900 
pounds,  with  a  deflection  of  at  least  jff  inch  before  rupture. 


WORKMANSHIP 

19.  General.  —  All  parts  forming  a  structure  shall  be  built  in  accord- 
ance with  approved  drawings.     The  workmanship  and  finish  shall  be  equal 
to  the  best  practice  in  modern  bridge  works. 

20.  Straightening  Material.  —  Material  shall  be  thoroughly  straightened 
in  the  shop,  by  methods  that  will  not  injure  it,  before  being  laid  off  or 
worked  in  any  way. 

21.  Finish.  —  Shearing  shall  be  neatly  and  accurately  done,  and  all 
portions  of  the  work  exposed  to  view  shall  be  neatly  finished. 

22.  Rivets.  —  The  size  of  rivets  called  for  in  the  plans  shall  be  under- 
stood to  mean  the  actual  size  of  the  cold  rivets  before  heating. 

23.  Rivet  Holes.  —  When  general  reaming  is  not  required,  the  diameter 
of  the  punch  for  material  not  over  f  inch  thick  shall  be  not  more  than 
Y1^  inch,  nor  that  of  the  die  more  than  \  inch,  larger  than  the  diameter  of 
the  rivet.     The  diameter  of  the  die  shall  not  exceed  the  diameter  of  the 
punch  by  more  than  one-fourth  the  thickness  of  the  material  punched. 
Material  over  f  inch  thick,  except  that  for  minor  details,  and  all  material 
where  general  reaming  is  required,  shall  be  sub-punched  and  reamed  as  per 
paragraphs  49,  50  and  51,  or  drilled  from  the  solid.    Holes  in  the  flanges  of 
rolled  beams  and  channels  used  in  the  floors  of  railroad  bridges  shall  be 


348  GRAPHICS  AND  STRUCTURAL  DESIGN 

drilled  from  the  solid.     Those  in  the  webs  of  same  shall  be  so  drilled  or  sub- 
punched  and  reamed. 

NOTE.  —  Mr.  Schneider  in  his  specifications  replaces  the  f  inch  occurring  in 
paragraph  23  by  f  inch. 

24.  Punching.  —  Punching  shall  be  accurately  done.     Slight  inaccuracy 
in  the  matching  of  holes  may  be  corrected  with  reamers.    Drifting  to 
enlarge  unfair  holes  will  not  be  allowed.     Poor  matching  of  holes  will  be 
cause  for  rejection  at  the  option  of  the  inspector. 

NOTE. —  Drifting  is  driving  a  taper  pin  through  holes  that  fail  to  match  properly; 
this  distorts  the  hole  and  injures  the  material. 

25.  Assembling.  —  Riveted  members  shall  have  all  parts  well  pinned 
up  and  firmly  drawn  together  with  bolts  before  riveting  is  commenced. 
Contact  surfaces  shall  be  painted  (see  paragraph  52). 

26.  Lattice  Bars.  —  Lattice  bars  shall  have  neatly  rounded  ends,  unless 
otherwise  called  for. 

27.  Web  Stiffeners.  —  Stiffeners  shall  fit  neatly  between  the  flanges  of 
girders.    Where  tight  fits  are  called  for,  the  ends  of  the  Stiffeners  shall  be 
faced  and  brought  to  a  true  contact  bearing  with  the  flange  angles. 

28.  Splice  Plates  and  Fillers.  —  Web  splice  plates,  and  fillers  under 
Stiffeners,  shall  be  cut  to  fit  within  |  inch  of  flange  angles. 

29.  Connection  Angles.  —  Connection  angles  for  floor  girders  shall  be 
flush  with  each  other  and  correct  as  to  position  and  length  of  girder.     In 
case  milling  is  required  after  riveting,  the  removal  of  more  than  TV  inch  from 
their  thickness  shall  be  cause  for  rejection. 

30.  Riveting.  —  Rivets    shall    be    driven    by   pressure    tools    wherever 
possible.     Pneumatic  hammers  shall  be  used  in  preference  to  hand  driving. 

31.  Rivets.  —  Rivets  shall  look  neat  and  finished,  with  heads  of  approved 
shape,  full  and  of  equal  size.     They  shall  be  central  on  shank  and  shall 
grip  the  assembled  pieces  firmly.      Recupping  and  calking  will  not  be 
allowed.    Loose,  burned  or  otherwise  defective  rivets  shall  be  cut  out  and 
replaced.    In  cutting  out  rivets  great  care  shall  be  taken  not  to  injure  the 
adjacent  metal.    If  necessary  they  shall  be  drilled  out. 

31  (a).  Heating  Rivets.  —  Rivets  shall  be  heated  to  a  light  cherry  red, 
in  a  gas  or  oil  furnace.  The  furnace  must  be  so  constructed  that  it  can 
be  adjusted  to  the  proper  temperature. 

NOTE.  —  Paragraph  31  (a)  is  inserted  by  Mr.  Schneider  but  is  not  generally  in- 
cluded by  others. 

32.  Field  Bolts.  —  Wherever  bolts  are  used  in  place  of  rivets  which 
transmit  shear,  the  holes  shall  be  reamed  parallel  and  the  bolts  turned  to 
a  driving  fit.    A  washer  not  less  than  \  inch  thick  shall  be  used  under  the 
nut. 


MATERIALS  349 

33.  Members  to  be  Straight.  —  The  several  pieces  forming  one  built-up 
member  shall  be  tight  and  fit  closely  together,  and  finished  members  shall 
be  free  from  twists,  bends  or  open  joints. 

34.  Finish  of  Joints.  —  Abutting  joints  shall  be  cut  or  dressed  true  and 
straight  and  fitted  closely  together,  especially  where  open  to  view. 

In  compression  joints,  depending  on  contact  bearing,  the  surfaces  shall 
be  truly  faced,  so  as  to  have  even  bearings  after  they  are  riveted  up  com- 
plete and  perfectly  aligned. 

35.  Field  Connections.  —  Holes  for  floor-girder  connections  shall  be  sub- 
punched  and  reamed  with  twist  drills  to  a  steel  template  i  inch  thick. 
Unless  otherwise  allowed,  all  other  field  connections  shall  be  assembled  in 
the  shop  and  the  unfair  holes  reamed;  when  so  reamed  the  pieces  shall  be 
match-marked  before  being  taken  apart. 

36.  Eyebars.  —  Eyebars  shall  be  straight  and  true  to  size,  and  shall  be 
free  from  twists,  folds  in  the  neck  or  head,  or  any  other  defect. 

Heads  shall  be  made  by  upsetting,  rolling  or  forging.  Welding  will  not 
be  allowed.  The  form  of  the  heads  will  be  determined  by  the  dies  in  use 
at  the  works  where  the  eyebars  are  made,  if  satisfactory  to  the  engineer, 
but  the  manufacturer  shall  guarantee  the  bars  to  break  in  the  body  with 
a  silky  fracture,  when  tested  to  rupture.  The  thickness  of  the  head  and 
neck  shall  not  vary  more  than  ^  inch  from  that  specified. 

37.  Boring   Eyebars.  —  Before   boring   each  eyebar  shall   be   properly 
annealed  and  carefully  straightened.     Pinholes  shall  be  in  the  center  line 
of  the  bars  and  in  the  center  of  the  heads.     Bars  of  the  same  length  shall 
be  bored  so  accurately  that,  when  placed  together,  pins  fa  inch  smaller 
in  diameter  than  the  pinholes  can  be  passed  through  the  holes  at  both  ends 
of  the  bars  at  the  same  time. 

38.  Pinholes.  —  Pinholes  shall  be  bored  true  to  gauges,  smooth  and 
straight,  at  right  angles  to  the  axis  of  the  member  and  parallel  to  each 
other,  unless  otherwise  called  for.     Wherever  possible,  the  boring  shall  be 
done  after  the  member  is  riveted  up. 

39.  Variation  in  Pinholes.  —  The  distance  center  to  center  of  pinholes 
shall  be  correct  within  fa  inch,  and  for  pins  up  to  5  inches  in  diameter  the 
diameter  of  the  hole  shall  not  exceed  the  diameter  of  the  pin  by  more  than 
fa  inch;  for  larger  pins  this  difference  shall  not  exceed  fa  inch. 

40.  Pins  and  Rollers.  —  Pins  and  rollers  shall  be  accurately  turned  to 
gauges  and  shall  be  straight,  smooth  and  entirely  free  from  flaws. 

41.  Pilot  Nuts.  —  At  least  one  pilot  and  driving  nut  shall  be  furnished 
for  each  size  of  pin  for  each  structure,  and  field  rivets  10  per  cent  in  excess 
of  the  number  of  each  size  actually  required. 

NOTE.  —  Pilot  and  driving  nuts  are  nuts  used  to  guide  and  protect  truss  pins 
during  driving  on  erection. 


350  GRAPHICS  AND  STRUCTURAL  DESIGN 

42.  Screw  Threads.  —  Screw  threads  shall  make  tight  fits  in  the  nuts 
and  shall  be  U.  S.  standard,  except  above  the  diameter  of  if  inches,  when 
they  shall  be  made  with  six  threads  per  inch. 

43.  Annealing.  —  Steel,  except  in  minor  details,  which  has  been  partially 
heated  shall  be  properly  annealed. 

44.  Steel  Castings.  —  All  steel  castings  shall  be  annealed. 

45.  Welds.  —  Welds  in  steel  will  not  be  allowed. 

46.  Bed  Plates.  —  Expansion  bed  plates  shall  be  planed  true  and  smooth. 
Cast  wall  plates  shall  be  planed  top  and  bottom;  the  cut  of  the  planing 
tool  shall  correspond  with  the  direction  of  expansion. 

47.  Shipping  Details.  —  Pins,  nuts,  bolts,  rivets  and  other  small  details 
shall  be  boxed  or  crated. 

ADDITIONAL  SPECIFICATIONS  WHEN  GENERAL  REAMING  AND  PLANING 

ARE  REQUIRED 

48.  Planing  Edges.  —  Sheared  edges  and  ends  shall  be  planed  off  at 
least  I  inch. 

49.  Reaming.  —  Punched  holes  shall  be  made  with  a  punch  T8g  inch 
smaller  in  diameter  than  the  nominal  size  of  the  rivets  and  shall  be  reamed 
to  a  finished  diameter  of  not  more  than  yV  inch  larger  than  the  rivet. 

50.  Reaming  after  Assembling.  —  Wherever  practicable,  reaming  shall 
be  done  after  the  pieces  forming  one  built  member  have  been  assembled  and 
firmly  bolted  together.     If  necessary  to  take  the  pieces  apart  for  shipping 
and  handling,  the  respective  pieces  reamed  together  shall  be  so  marked  that 
they  may  be  reassembled  in  the  same  position  in  the  final  setting  up.    No 
interchange  of  reamed  parts  will  be  allowed. 

51.  Removing  Burrs.  —  The  burrs  on  all  reamed  holes  shall  be  removed 
by  a  tool  countersinking  about  yV  inch. 

PAINTING 

51  (a).  Shop  Painting.  —  Steel  work,  before  leaving  the  shop,  shall  be 
thoroughly  cleaned  and  given  one  good  coating  of  pure  linseed  oil,  or  such 
paint  as  may  be  called  for,  well  worked  into  all  joints  and  open  spaces. 

52.  In  riveted  work,  the  surfaces  coming  in  contact  shall  each  be  painted 
before  being  riveted  together. 

53.  Pieces  and  parts  which  are  not  accessible  for  painting  after  erection, 
including  tops  of  stringers,  eyebar  heads,  ends  of  posts  and  chords,  etc., 
shall  be  given  two  coats  of  paint  before  leaving  the  shop. 

54.  Steel  work  to  be  entirely  embedded  in  concrete  shall  not  be  painted. 

55.  Painting  shall  be  done  only  when  the  surface  of  the  metal  is  per- 
fectly dry.     It  shall  not  be  done  in  wet  or  freezing  weather,  unless  protected 
under  cover. 


MATERIALS  351 

56.  Machine-finished   Surfaces.  —  These  shall  be   coated   with  white 
lead  and  tallow  before  shipment  or  before  being  put  out  into  the  open  air. 

57.  Field  Painting.  —  After  the  structure  is  erected,  the  metal  work  shall 
be  painted  thoroughly  and  evenly  with  an  additional  coat  of  paint,  mixed 
with  pure  linseed  oil,  of  such  quality  and  color  as  may  be  selected. 

Succeeding  coats  of  paint  shall  vary  somewhat  in  color,  in  order  that 
there  may  be  no  confusion  as  to  the  surfaces  that  have  been  painted. 

INSPECTION  AND  TESTING 

58.  Facilities    for    Inspection.  —  The    manufacturer    shall    furnish    all 
facilities  for  inspecting  and  testing  the  weight,  quality  of  material  and 
workmanship  at  the  shop  where  the  material  is  manufactured.     He  shall, 
if  required,  furnish  a  suitable  testing  machine  for  testing  full-sized  members. 

The  manufacturer  shall  prepare  all  test  pieces  for  the  machine,  free 
of  cost. 

59.  Access  to  Shop.  —  When  an  inspector  is  furnished  by  the  purchaser, 
he  shall  have  full  access,  at  all  times,  to  all  parts  of  the  shop  where  material 
under  his  inspection  is  being  manufactured. 

60.  The  purchaser  shall  be  furnished  complete  shop  plans,  and  must  be 
notified  well  in  advance  of  the  start  of  work  in  the  shop,  in  order  that  he 
may  have  an  inspector  on  hand  to  inspect  material  and  workmanship. 

61.  Shipping  Invoices.  —  Complete  copies  of  shipping  invoices  shall  be 
furnished  to  the  purchaser  with  each  shipment. 

62.  Mill   Orders.  —  The  purchaser  shall  be  furnished  with  complete 
copies  of  mill  orders,  and  no  material  shall  be  rolled  and  no  work  done  before 
he  has  been  notified  as  to  where  the  orders  have  been  placed,  so  that  he  may 
arrange  for  the  inspection. 

63.  Inspector's  Mark.  —  The  inspector  shall  stamp  with  a  private  mark 
each  piece  accepted.     Any  piece  not   so  marked  may  be  rejected  at  any 
tune,  at  any  stage  of  the  work.     If  the  inspector,  through  an  oversight  or 
otherwise,  has  accepted  material  or  work  which  is  defective  or  contrary  to 
specifications,  this  material,  no  matter  hi  what  stage  of  completion,  may  be 
rejected  by  the  purchaser. 

FULL-SIZED  TESTS 

64.  Full-sized  Tests.  —  Full-sized  parts  of  the  structure  may  be  tested 
at  the  option  of  the  purchaser.     Such  tests  on  eyebars  and  similar  members, 
to  prove  the  workmanship,  shall  be  made  at  the  manufacturer's  expense, 
and  shall  be  paid  for  by  the  purchaser,  at  contract  price,  if  the  tests  are 
satisfactory.     If  the  tests  are  not  satisfactory,  the  members  represented  by 
them  will  be  rejected.     The  expense  of  testing  members,  to  prove  their 
design,  shall  be  paid  for  by  the  purchaser. 


352  GRAPHICS  AND   STRUCTURAL  DESIGN 

65.  Eyebar  Tests.  —  In  eyebar  tests,  the  minimum  ultimate  strength 
shall  be  55,000  pounds  per  square  inch.     The  elongation  in  10  feet,  in- 
cluding fracture,  shall  be  not  less  than  15  per  cent.     Bars  shall  break  in 
the  body  and  the  fracture  shall  be  silky  or  fine  granular.     The  elastic  limit 
as  indicated  by  the  drop  of  the  mercury  shall  be  recorded.     Should  a  bar 
break  in  the  head  and  develop  the  specified  elongation,  ultimate  strength 
and  character  of  fracture,  it  shall  not  be  cause  for  rejection,  provided  not 
more  than  one-third  of  the  total  number  of  bars  break  in  the  head. 

SPECIFICATIONS   FOR   STEEL   MILL   BUILDINGS 

66.  Dimensions.  —  By  height  shall  be  understood  the  distance  from  the 
under  side  of  the  lower  chord  of  the  roof  truss  to  the  tops  of  the  foundations. 

The  width  and  length  of  the  building  shall  be  measured  to  the  outsides 
of  the  framing  or  sheathing. 

67.  Spans.  —  The  spans  of  trusses,  beams  and  girders  for  the  purpose  of 
calculations  shall  be  assumed  as  the  distance  center  to  center  of  bearings 
for  supported  spans,  while  they  will  be  considered  as  the  distance  back  to 
back  of  framing  angles  when  the  trusses,  beams  or  girders  are  framed  into 
columns. 

LOADS 

68.  Dead  Loads.  —  The  dead  load  is  the  weight  of  all  permanent  con- 
struction and  fixtures.    The  calculated  weights  shall  be  based  upon  those 
hereafter  given. 

(a)    Weight  of  Trusses.  —  The  weight  of  a  truss  may  be  estimated  as 


w  = 

where 

W  =  weight  of  truss  per  square  foot  of  building  area, 

L  =  span  of  truss,  in  feet, 

D  =  distance  center  to  center  of  trusses,  in  feet, 

P  —  load  per  square  foot  on  the  truss. 

(b)   Weight  of  Purlins.  —  The    weight  of  purlins  per  square  foot  of 
horizontal  projection  of  the  roof  may  be  taken  as 

Wi=  VpixD-l, 
45  4 

where 

Wi  =  weight  of  purlins  per  square  foot  of  building, 
D  =  distance  center  to  center  of  the  trusses, 
PI  =  load  per  square  foot  on  purlins. 


MATERIALS  353 

(c)  Weight  of  Roof   Coverings.  —  The  weights  of  roof  coverings  not 
including  sheathing  per  square  foot  of  actual  roof  surface  will  average 

Tin i       Ib.     Corrugated  steel,  No.  18 3.0  Ibs. 

Slate,  ^-inch 6.6  Ibs.  Corrugated  steel,  No.  20.  ...  2.3  Ibs. 

Slate,  |-inch 4.4  Ibs.     Felt  and  gravel 7-9  Ibs. 

Terra-cotta,  i  inch  thick. .  6.0  Ibs. 

(d)  Weight  of  Sheathing.  - 

Wooden  sheathing,  per  foot,  board  measure 3.5  ibs. 

Concrete,  i  inch  thick,  per  square  foot 10-12  Ibs. 

NOTE.  —  A  foot  board  measure  is  a  piece  12  inches  square  and  i  inch  thick. 

LIVE  LOADS 

69.  (a)  Live  Loads.  —  These  shall  not  be  less  than  the  following  uniform 
loads : 

Warehouses 1 20  pounds  per  square  foot. 

Foundry  charging  floors 300  pounds  per  square  foot. 

Power  houses,  uncovered  floors .  .  .  200  pounds  per  square  foot. 

In  all  cases  special  concentrations,  such  as  engines,  turbines,  boilers,  chim- 
neys, etc.,  should  be  considered  from  actual  weights. 

(b)  Crane  Loads.  —  Where  available  the  actual  weights  and  dimensions  of 
traveling  cranes  should  be  used;  otherwise  the  table  of  weights  given  on  page 
354  and  taken  from  the  Specifications  of  Mr.  C.  C.  Schneider  may  be  used. 

(c)  Distribution  of  Wheel  Loads.  —  Wheel  loads  transferred  to  beams  or 
girders  through  rails  may  be  considered  as  distributed  a  distance  equal  to 
the  depth  of  the  girder  or  beam  but  not  exceeding  30  inches.     (Ostrup.) 

(d)  Lateral  Loading.  —  Besides  the  vertical  loading,  beams  and  girders 
carrying  traveling  cranes  shall  be  designed  so  that  their  upper  flanges  shall 
in  addition  resist  a  lateral  bending  due  to  one-twentieth  of  the  crane's 
capacity  acting  at  each  bridge-truck  wheel. 

(e)  Impact.  —  An  addition  of  25  per  cent  of  the  live-load  bending  mo- 
ments and  shears  shall  be  made  to  all  beams,  girders  and  columns  carrying 
traveling  cranes. 

70.  Flat  Roofs.  —  Flat  roofs  liable  to  be  loaded  with  people  should  be 
designed  as  floors.     This  would  subject  them  to  a  possible  additional  load 
of  40  pounds  per  square  foot. 

71.  Wind  Loads.  —  The  normal  wind  pressure  on  a  roof  whose  angle  of 
slope  does  not  exceed  45  degrees  shall  be  based  upon  a  horizontal  wind 
pressure  of  30  pounds  per  square  foot  and  its  magnitude  shall  be  estimated 
by  the  formula, 

W       Wh-A 

rV  n  —  > 

45 


354 


GRAPHICS  AND   STRUCTURAL  DESIGN 


where  WH  is  the  horizontal  wind  pressure  in  pounds  per  square  foot,  A  is 
the  angle  of  inclination  of  the  roof  in  degrees  and  Wn  is  the  resulting  normal 
wind  pressure  on  the  roof  in  pounds  per  square  foot. 

NOTE.  —  Some  designers  assume  a  horizontal  wind  pressure  of  from  20  to  30 
pounds  per  square  foot  acting  on  the  vertical  projection  of  the  roof. 

WEIGHTS  AND  DIMENSIONS  OF  TYPICAL  TRAVELING 

CRANES 


Capacity. 

Span. 

Wheel  base. 

Maximum 
wheel  load. 

Side 
clearance. 

Vertical 
clearance. 

Weight  of  rail  for 

Tons. 

Feet. 

Ft.  Ins. 

Pounds. 

Inches. 

Ft.  Ins. 

Runway 
girder. 

Beams. 

5 
5 

40 
60 

8      6 
9     0 

12,000 
13,000 

IO 
IO 

6     0 
6     0 

40 
40 

40 
40 

10 
10 

40 
60 

9     ? 
9     6 

19,000 
2I,OOO 

IO 
IO 

6     0 
6     o 

45 
45 

40 
40 

15 
15 

40 
60 

9     6 
10     o 

25,000 
29,000 

IO 
10 

7     o 
7     o 

5° 
50 

50 
50 

20 
20 

40 
60 

IO      O 
IO      O 

33,000 
36,000 

12 
12 

7     o 
7     o 

55 
55 

50 
50 

25 
25 

40 
60 

IO      0 

10     6 

40,000 
44,000 

12 
12 

8     o 
8     o 

60 
60 

50 
50 

30 
30 

40 
60 

10     6 

II       0 

48,000 
52,000 

12 
12 

8     o 
8     o 

70 
70 

60 
60 

40 
40 

40 
60 

II       0 
12       O 

64,000 
70,000 

12 
12 

9     o 
9     o 

80 
80 

60 
60 

50 
50 

40 
60 

II       0 
12       0 

72,000 
80,000 

14 
14 

9     o 
9     o 

IOO 
100 

60 
60 

NOTE.  —  The  rail  weights  are  in  pounds  per  yard.     The  side  clearance  is  from 
the  center  of  the  rail,  while  the  vertical  clearance  is  from  the  top  of  the  rail. 


72.  Wind  Pressure  on  the  Sides  and  Ends  of  the  Building.  —  In  build- 
ings not  exceeding  30  feet  to  the  eaves  this  pressure  may  be  taken  at  20 
pounds  per  square  foot,  while  it  should  be  increased  up  to  30  pounds  for 
buildings  60  feet  or  more  to  the  eaves. 

73.  Wind  Pressure  on  Framework.  —  The  wind  pressure  shall  be  esti- 
mated for  the  wind  acting  horizontally  in  any  direction  upon  the  total 
exposed  surface  of   the  framework.     Such  pressure  shall  be  assumed  at 


MATERIALS  355 

30  pounds  per  square  foot  of  exposed  surface.     (This  applies  during  con- 
struction.) 

74.  Snow  Loads.  —  For  latitudes  of  about  40  degrees  the  snow  load 
may  be  assumed  at  25  pounds  per  square  foot  of  horizontal  projection  of  the 
roof  for  flat  roofs  and  those  inclined  up  to  angles  of  15  degrees;  above  this 
slope  the  load  may  be  decreased  uniformly  so  that  at  45  degrees  it  would 
become  zero.     (The  slope  corresponding  to  15  degrees  is  about  one  in  four.) 

These  loads  should  be  increased  for  higher  latitudes  and  reduced  for 
lower  latitudes.  In  tropical  countries  snow  loads  may  be  neglected. 

The  amount  and  character  of  the  precipitation  in  any  locality  may  also 
affect  the  loading  which  would  naturally  be  lower  in  arid  sections. 

NOTE.  —  According  to  Mr.  C.  C.  Schneider  in  climates  corresponding  to  that  of 
New  York,  ordinary  roofs  up  to  spans  of  80  feet  may  be  designed  for  the  following 
minimum  equivalent  loads  per  square  foot  of  horizontal  projection  of  the  roof. 
These  loads  then  replace  the  dead  loads,  wind  loads  and  snow  loads  given  above. 

Lbs. 

Gravel  or          fOn  boards,  slope  i  to  6,  or  less 50 

composition  -4  On  boards,  slope  exceeding  i  to  6 45 

roofing  [  On  3-inch  flat  tile  or  cinder  concrete 60 

Corrugated  sheets,  on  boards  or  purlins 40 

Slate,  on  boards  or  purlins 50 

Slate,  on  3-inch  flat  tile  or  cinder  concrete 65 

Tile  on  steel  purlins 55 

Glass 45 

Where  no  snow  is  to  be  expected  the  above  loads  may  be  reduced  10  pounds, 
excepting  that  no  roof  should  be  designed  for  a  load  less  than  40  pounds  per 
square  foot. 

75.  Minimum  Loads  on  Purlins  and  Roof  Coverings.  —  Purlins  and  roof 
coverings  shall  not  be  designed  for  normal  loads  under  30  pounds  per  square 
foot. 

76.  Loads  on  Foundations.  —  The  areas  of  foundation  piers  shall  be 
proportional  to  their  respective  dead  loads;   in  no  case,  however,  shall  the 
combined  live  and  dead  loads  on  a  pier  divided  by  its  base  area  exceed  the 
permissible  pressure  on  the  soil. 

NOTE.  —  The  desire  is  to  obtain  uniform  settlement.  In  making  the  foundations 
proportional  to  the  dead  loads  it  is  considered  that  as  the  dead  loads  act  continu- 
ously the  settlement  will  depend  more  on  dead  loads  than  on  the  possibly  very 
intermittent  action  of  the  live  loads. 


356  GRAPHICS  AND   STRUCTURAL  DESIGN 

UNIT  STRESSES   AND   PROPORTION   OF   PARTS 

SUBSTRUCTURE 

77.  Pressures  on  Soils.  —  The  pressures  on  the  soil  at  the  base  of  the 
foundation  shall  not  exceed  the  following  in  tons  of  2000  pounds. 

Clay,  soft i 

Ordinary  clay,  or  dry  sand  mixed  with  clay 2 

Dry  sand  and  dry  clay 3 

Hard  clay  and  firm  coarse  sand 4 

Firm  gravel  and  coarse  sand •        6 

Rock,  according  to  condition 15-200 

78.  Compressive   Stresses   in   Masonry.  —  The   following   compressive 
stresses  will  be  permitted  in  masonry  structures: 

Lbs.  per  sq.  in. 

Common  brick,  in  Portland  cement  mortar 170 

Hard-burned  brick,  in  Portland  cement  mortar 200 

Rubble  masonry,  in  Portland  cement  mortar 150 

Sandstone  masonry,  first  class 280 

Limestone  masonry,  first  class 350 

Granite  masonry,  first  class 420 

Portland  cement  concrete,  1-2-4 400 

Portland  cement  concrete,  1-2-5 300 

79.  Wall-plate    Pressure.  —  According   to   Mr.    C.    C.   Schneider,   the 
pressure  of  beams,  girders,  wall  plates,  columns,  etc.,  on  masonry  shall  not 
exceed  the  following : 

Lbs.  per  sq.  in. 

On  brick  work  in  cement  mortar 300 

On  rubble  masonry  in  cement  mortar 250 

On  Portland  cement  concrete 600 

On  first-class  sandstone,  dimension  stone 400 

On  first-class  limestone 500 

On  first-class  granite 600 

80.  Bearing  Power  of  Piles.  —  Piles  shall  not  be  spaced  closer  than  30 
inches  center  to  center.    The  maximum  load  on  any  pile  shall  not  exceed 
40,000  pounds,  or  600  pounds  per  square  inch  of  average  cross  section. 
When  driven  to  rock  or  equivalent  bearing  through  loose  or  wet  soil,  which 
gives  them  no  lateral  support,  the  limiting  load  shall  be  determined  by 
reducing  600  pounds,  the  maximum  allowable  compression,  by  a  suitable 
column  formula. 

81.  Walls  shall  be  built  in  accordance  with  the  local  "  building  code  " 
when  that  is  available,  otherwise  they  may  be  made  the  thicknesses  given 
on  page  324. 


MATERIALS 


357 


82.  Pillars.  —  (a)  When  concentrically  loaded  and  having  a  height  not 
exceeding  twelve  times  their  least  dimension,  pillars  may  be  loaded  until 
the  fiber  stresses  reach  the  figures  given  in  paragraph  78. 

(b)  When  eccentrically  loaded  the  resultant  pressure  must  pass  within 
the  kern  of  the  section  and  the  maximum  pressure  should  not  exceed  that 
given  in  paragraph  78. 

NOTE.  —  For  the  explanation  of  kern  of  a  section  see  page  238. 

UNIT  STRESSES  IN  STEEL  WORK 

83.  Permissible   Stresses.  —  The  resulting  stresses  due  to  dead  load, 
snow  load,  wind  load  and  impact  shall  not  exceed  the  following  limiting 
values,  excepting  where  permitted  in  accordance  with  paragraph  91. 

84.  Tension,  net  section,  rolled  steel,  16,000  pounds  per  square  inch. 

85.  Direct  Compression,  rolled  steel  and  steel  castings,  16,000  pounds 
per  square  inch. 

86.  Bending  Stresses,  on  extreme  fibers  of  rolled  shapes,  built  sections, 
girders  and  steel  castings,  net  section,  16,000  pounds  per  square  inch. 

On  extreme  fibers  of  pins,  24,000  pounds  per  square  inch. 

87.  Shearing  Stresses.  —  Lbs.  per  sq.  in. 

On  shop  rivets  and  pins 12,000 

On  bolts  and  field  rivets 10,000 

On  plate  girders  and  beams,  cross  section,  average 10,000 

88.  Bearing  Stresses.  —  Lbs.  per  sq.  in. 

On  shop  rivets  and  pins 24,000 

On  field  rivets  and  pins 20,000 

89.  Pressure  on  Expansion  Rollers.  —  The  pressure  per  linear  inch  on 
expansion  rollers  shall  not  exceed  600  pounds  per  inch  of  roller  diameter. 

90.  Axial  Compression  in  Columns.  —  The  load  per  square  inch  of  gross 

section   for   columns   axially  loaded  shall   neither  exceed  16,000— (C-) 

pounds,  nor  14,000  pounds.  Here  /  =  the  length  of  the  member  in  inches; 
r  —  the  corresponding  radius  of  gyration  of  the  section  in  inches;  and  C 
has  the  following  values: 


Condition  of  ends. 

C. 

Both  ends  hinged  or  butting  

70 

Both  ends  fixed.          

2<? 

One  end  fixed,  the  other  hinged  

47 

One  end  fixed  the  other  free 

I  dO 

358 


GRAPHICS  AND  STRUCTURAL  DESIGN 


NOTE.  —  The  specifications  issued  Dec.  i,  1912,  by  the  American  Bridge  Co. 
permit  loads  upon  columns,  concentrically  loaded,  as  given  by  the  formula  19,000  — 

(  zoo  X  -  1  but  not  to  exceed  13,000  pounds  per  square  inch  and  applying  to  values 
of  -  up  to  1 20. 

For  values  of  -  from  120  to  200  they  give  the  following  table: 


I 

r 

Allowable  load, 
Ibs.  per  sq.  in. 

I 
r 

Allowable  load, 
Ibs.  per  sq.  in. 

120 
130 
140 

7000 
6500 
6000 

160 
170 
180 

5000 
4500 
4000 

ISO 

5500 

190 

3500 

200 

3000 

91.  Stresses  in  Bracing  and  Members  with  Combined  Stresses.  —  In 
bracing  and  members  subjected  to  combined  stresses  resulting  from  wind 
and  other  loadings  the  permissible  working  stresses  previously  given  may 
be  increased  25  per  cent. 

92.  Eccentric  Loading.  —  In  designing  columns  provision  must  be  made 
for  eccentric  loading. 

93.  Combined  Stresses.  —  Members  subjected  to  combined  axial  and 
bending  stresses  shall  be  designed  so  that  the  greatest  fiber  stress  due  to 
the  combined  stresses  shall  not  exceed  the  fiber  stress  permitted  in  the 
member. 

94.  Alternate  Stresses.  —  Members  subjected  to  a  reversal  of  the  stresses 
shall  be  proportioned  for  that  stress  giving  the  greater  section.     The  con- 
nections shall  be  designed  to  carry  the  sum  of  the  stresses. 

95.  Net  Sections.  —  In  finding  net  sections,  the  rivet  holes  shall  be 
assumed  as  \  inch  larger  than  the  nominal  diameter  of  the  rivet. 

96.  Limiting  Lengths  of  Columns.  —  Columns  assumed  as  having  hinged 
or  butting  ends  shall  have  unsupported  lengths  not  exceeding  125  times  their 
least  radius  of  gyration  for  main  members  nor  more  than  150  times  their 
least  radius  of  gyration  for  wind  bracing  or  secondary  members. 

NOTE.  —  The  recent  specifications  of  the  American  Bridge  Co.  place  these  lengths 
at  1 20  and  200  times  the  least  radius  of  gyration. 

97.  Limiting  Lengths  of  Tension  Pieces.  —  Riveted  tension  pieces  shall 
have  lengths  not  exceeding  200  times  their  radii  of  gyration  about  their 
horizontal  axes,  when  used  in  horizontal  or  inclined  positions.     The  length 
for  this  calculation  shall  be  considered  that  of  its  horizontal  projection. 


MATERIALS  359 

98.  Rolled  Sections  used  as  Beams.  —  Rolled  sections  used  as  beams 
shall  be  proportioned  by  their  moments  of  inertia. 

99.  Plate  Girders.  —  Plate  girders  shall  be  designed  upon  the  assump- 
tion that  one-eighth  of  the  web  acts  as  flange  area. 

NOTE.  —  For  buildings  and  similar  structures  it  is  frequently  specified  that  the 
bending  stresses  shall  be  resisted  by  the  flanges  and  that  the  web  shall  resist  shear 
only. 

NOTE.  —  The  recent  specifications  of  the  American  Bridge  Co.  permit  the  design 
of  girders  by  their  moments  of  inertia,  thus  making  paragraph  99  apply  to  both 
girders  and  rolled  sections. 

100.  Web  Thickness  of  Plate  Girders.  —  The  web  thickness  shall  be  not 
less  than  T^  of  the  unsupported  distance  between  flange  angles. 

101.  Web  Stiff eners.  —  Web  stiff eners  shall  be  placed  on  both  sides  of 
the  web,  with  a  close  bearing  against  upper  and  lower  flange  angles,  at  the 
ends  and  inner,  edges  of  bearing  plates,  at  all  local  and  concentrated  loads; 
where  the  thickness  of  the  web  plate  is  less  than  fa  of  the  unsupported 
distance  between  flange  angles,  stiffeners  shall  be  placed  at  intervals  along 
the  girder  about  equal  to  the  girder  depth  but  not  exceeding  5  feet. 

NOTE.  —  The  recent  specifications  of  the  American  Bridge  Co.  place  this  limiting 
distance  at  6  feet. 

102.  Compression  Flange.  —  The  area  of  the  compression  flange  must 
at  least  equal  the  area  of  the  tension  flange  at  the  same  section  of  the  girder. 
Where  the  unsupported  distance  along  the  girder  exceeds  20  times  the  width 
of  the  flange  the  fiber  stress  in  the  compression  flange  must  be  that  deter- 
mined by  considering  the  flange  as  a  column  and  buckling  laterally. 

NOTE.  —  The  practice  on  this  point  varies  considerably  among  designers.  Mr. 
C.  C.  Schneider  limits  the  fiber  stress  in  the  compression  flanges  at  all  times  to 
that  given  by  the  following  formulae.  When  the  flange  section  consists  of  angles 

and  plates  the  fiber  stress  is  limited  by  16,000  —  f  200  X  r  ) ,  and  where  the  flange 

is  a  channel  section  by  16,000  —  (  150  X   T  )•     Here  /  is  the  distance  along  the 

flange  between  points  of  support  and  b  is  the  width  of  the  flange  plate,  both  being 
in  inches. 

The  requirements  as  stated  by  the  specifications  of  the  American  Bridge  Co.  are 
that  the  flange  must  be  braced  at  intervals  not  exceeding  40  times  the  width  of 
the  flange  plates,  and  that  the  limiting  fiber  stress  for  unsupported  flange  lengths 
exceeding  10  times  the  widths  of  the  flanges  shall  be  that  given  by  19,000  — 

( 300  XT).     The  symbols  have  the  same  significance  as  those  given  above. 

Other  specifications  permit  an  unsupported  flange  length  of  12,  16  and  even 
30  times  the  width  of  -the  flange  plates  before  requiring  any  reduction  in  the  fiber 
stress. 


360 


GRAPHICS  AND  STRUCTURAL  DESIGN 


103.  Depths  of  Rolled  Beams.  —  The  depths  of  rolled  beams  where  used 
for  floors  shall  be  not  less  than  ^  of  the  span;  when  used  for  purlins  they 
may  be  as  small  as  fa  of  the  span.     Floor  beams  when  subjected  to  con- 
siderable vibration  or  shock  should  be  made  not  less  than  ^~  of  their  span. 

NOTE.  —  The  purpose  of  this  requirement  is  to  limit  the  deflection. 

104.  Permissible  Stresses  on  Cast  Iron.  — 

Tension 2,500  Ibs.  per  sq.  in. 

Compression 12,000  Ibs.  per  sq.  in. 

Shear J^oo  Ibs.  per  sq.  in. 

105.  Permissible  Stresses  on  Timber.  —  The  timber  in  the  structure 
may  be  designed  with  the  following  fiber  stresses  given  in  pounds  per  square 
inch. 


Columns  under 

10  diameters. 

Bending 

Bearing 

Shear 
along 

Weight 
per 

stress. 

on  end. 

Com- 

Bearing 

fibers. 

cu.  ft. 

pres- 

across 

sion. 

fibers. 

White  oak 

I2OO 

1400 

IOOO 

TOO 

2OO 

PQ 

Long-leaf  yellow  pine  . 

I=COO 

I^OO 

IOOO 

JW 

•2^0 

IOO 

o(J 
*8 

White  pine  and  spruce  

-LO'~"ta/ 

900 

A  ^  W 
QOO 

700 

oo^ 
200 

IOO 

o° 

24 

Hemlock  ....       

700 

700 

500 

200 

IOO 

25 

1  06.   Timber  Columns.  —  The  allowable  fiber  stress  on  columns  exceed- 
ing 10  diameters  in  length  may  be  obtained  by  the  formula 


Vioo  X 

where  C  is  the  unit  stress  given  above  for  short  columns;  /  is  the  length 
and  b  the  least  width  of  the  column,  both  in  inches. 


DETAILS  OF  CONSTRUCTION 

107.  General.  —  Adjustable  members  will  be,  if  possible,  avoided  in  all 
parts  of  the  structure. 

1 08.  Symmetrical   Sections.  —  All   sections   shall  preferably  be  made 
symmetrical. 

109.  Minimum  Number  of  Rivets.  —  Exceptirig  lattice  bars  all  connec- 
tions shall  have  at  least  two  rivets. 

no.   Minimum  Thickness  of  Material.  —  Excepting  for  lining  or  fillers 
no  material  shall  be  less  than  \  inch  thick. 


MATERIALS  361 

NOTE.  —  Some  designers  place  this  minimum  thickness  at  f\  inch  for  material 
protected  from  the  weather  and  not  affected  by  injurious  gases.  Where  the  con- 
ditions might  be  expected  to  cause  rapid  deterioration  of  the  material  it  is  fre- 
quently made  in  excess  of  the  |  inch  first  stated. 

in.  Connections.  —  Connections  shall  be  made  strong  enough  to 
develop  the  full  strength  of  the  members. 

112.  Floor  Beams.  —  Floor  beams  shall  ordinarily  be  rolled  sections. 

113.  Trusses.  —  Trusses  shall  preferably  be  riveted  structures.     Heavy 
trusses  of  long  span  may  be  designed  as  pin-connected  structures. 

Roof  trusses  shall  preferably  permit  of  placing  purlins  at  panel  points 
only.  Where  this  would  prove  uneconomical  the  upper  chord  may  be 
designed  to  resist  combined  flexure  and  compression  and  the  purlins  then 
placed  where  convenient. 

114.  Bracing.  —  Lateral,    longitudinal    and    transverse    bracing    in    all 
structures  is  preferably  composed  of  rigid  members,  and  should  be  designed 
of  sufficient  strength  to  withstand  the  wind  pressure  and  any  other  forces 
acting  on  it  both  during  and  after  erection. 

Trusses  should  be  braced  in  pairs  in  the  plane  of  their  lower  and  upper 
chords. 

115.  Column  Splices.  —  Column  splices  should  be  designed  to  resist  the 
bending  as  well  as  the  direct  stresses. 

116.  Intersecting  Members.  —  Intersecting  members  should  be  located 
so  that  the  axes  through  their  centers  of  gravity  shall  intersect  in  a  point. 

117.  Separators.  —  When  rolled  beams  are  placed  side  by  side  to  act 
as  a  single  girder  they  shall  be  secured  together  by  bolts  and  separators 
at  intervals  not  exceeding  5  feet.     Beams  of  greater  depth  than  10  inches 
should  have  two  bolts  in  each  separator.     Parallel  beams  carrying  arched 
floors,  whether  of  the  ordinary  or  flat  arch  type,  shall  have  tie  bolts  spaced 
not  farther  apart  than  8  times  the  depth  of  the  beams.     These  tie  bolts 
should  be  placed  well  below  the  centers  of  the  beams. 

118.  Width  of  Flange  Plates.  —  Flange  plates  shall  be  limited  in  width 
so  as  not  to  extend  more  than  6  inches  beyond  the  line  of  rivets  securing 
the  flange  plates  to  the  flange  angles,  nor  should  this  dimension  exceed  8 
tunes  the  thickness  of  the  thinnest  flange  plate. 

119.  Web  Splices.  —  All  web  splices  must  have  a  plate  on  each  side  of 
the  web,  and  these  plates  must  be  able  to  transmit  through  their  rivets  the 
full  stress  coming  on  the  splice  at  that  section. 

RIVETING 

1 20.  Riveting.  —  The  minimum  distance  between  centers  of  rivet  holes 
shall  be  three  diameters  of  the  rivet;    it  is  preferred,  however,  that  this 
distance  should  be  not  less  than  3  inches  for  f-inch  rivets,  2\  inches  for 
1-inch  rivets,  2  inches  for  f-inch  rivets,  and  if  inches  for  £-mch  rivets. 


362  GRAPHICS  AND   STRUCTURAL  DESIGN 

The  maximum  pitch  in  the  line  of  the  stress  for  members  composed  of 
plates  and  shapes  will  be  6  inches  for  £-inch  and  f-inch  rivets,  4!  inches  for 
f-inch  rivets  and  4  inches  for  ^-inch  rivets. 

121.  Rivet  Spacing  in  Angles.  —  For  angles  built  into  sections  with  two 
gauge  lines,  the  rivets  being  staggered,  the  maximum  pitch  in  each  line  shall 
be  twice  that  given  in  paragraph  1 20. 

122.  Riveting  Plates.  —  Where  two  or  more  plates  are  in  contact,  they 
shall  be  held  together  by  rivets  whose  distance  apart  in  either  direction  does 
not  exceed  12  inches. 

The  pitch  of  rivets  in  the  direction  of  the  stress  shall  not  exceed  6  inches, 
nor  1 6  times  the  thickness  of  the  thinnest  outside  plate. 

The  spacing  at  right  angles  to  the  stress  shall  not  exceed  50  times  the 
thickness  of  the  thinnest  plate.  (Some  specifications  make  this  40  instead 
of  50.) 

123.  Rivet  Spacing  from  Edges.  —  The  minimum  distance  of  a  rivet 
from  a  sheared  edge  shall  be  1 1  inches  for  f-inch  rivets,  i  i  inches  for  f-inch 
rivets,  1 1  inches  for  f-inch  rivets,  and  i  inch  for  £-inch  rivets.     The  mini- 
mum distances  from  rolled  edges  shall  be  i|,  i|,  i  and  £  inch,  respectively. 

The  maximum  distance  of  a  rivet  from  an  edge  shall  be  8  times  the  thick- 
ness of  a  plate. 

124.  Calculations  of  Rivet  Strength.  —  All  calculations  of  rivet  values 
both  in  shear  and  in  bearing  shall  be  based  upon  their  nominal  diameters. 

The  grips  of  rivets  should  preferably  not  exceed  4  diameters  of  the  rivets. 

NOTE.  —  The  grip  is  the  total  thickness  of  the  pieces  held  by  the  rivet.  This 
limitation  is  made  to  insure  the  rivet  completely  filling  the  rivet  hole  when  driven. 

125.  Maximum  Rivet  Diameters  in  Angles.  —  In  main  members  the 
diameters  of  rivets  in  angles  shall  not  exceed  one-quarter  the  widths  of  the 
legs  in  which  they  are  driven.     In  other  places  these  diameters  may  be 
increased  |  inch. 

126.  Pitch  of  Rivets  at  Ends  of  Compression  Pieces.  —  The  pitch  of 
rivets  at  the  ends  of  compression  pieces  shall  not  exceed  4  times  the  rivet 
diameter  for   a  distance  equal  to  i|  times  the  maximum  width  of   the 
member. 

127.  Tie  Plates.  —  Latticed  sides  of  compression  members  shall  have  tie 
plates  as  close  to  the  ends  as  practicable,  and  at  all  intermediate  points  at 
which  the  lacing  may  be  omitted.     The  lengths  of  these  tie  plates  shall  at 
least  equal  the  distance  between  the  rivet  lines  securing  them  to  the  flanges. 
The  intermediate  plates  may  be  but  one-half  this  length.     The  thickness 
of  these  tie  plates  must  be  at  least  -fa  of  the  distance  between  the  lines  of 
rivets  securing  the  plate  to  the  flanges. 


MATERIALS  363 

128.  Lattice  Bars.  —  Lattice  bars  shall  be  proportioned  to  resist  the 
shear  corresponding  to  the  allowance  for  bending  made  in  the  column 
formula,  paragraph  90.     The  minimum  thickness  of  lattice  bars  shall  be 
?V  of  their  lengths  for  single  lattice  bars  and  ^V  of  their  lengths  for  double 
lattice  bars  riveted  at  their  centers. 

The  minimum  widths  of  lattice  bars  shall  be  as  follows: 

For  i5-inch  channels,  or  built  sections 

with  35  and  4-inch  angles,  2\  inches  (I-inch  rivets). 

For  12,  10  and  Q-inch  channels,  or  built 

sections  with  3-inch  angles,  2^  inches  (f-inch  rivets). 

For   8   and    7-inch  channels,  or  built 

sections  with  2|-inch  angles,  2  inches  (|-inch  rivets). 

For   6  and   5-inch  channels,  or   built 

sections  with  2-inch  angles,  if  inches  (i-inch  rivets). 

NOTE.  —  The  specifications  of  the  American  Bridge  Co.  require  the  lattice  bars 
to  be  designed  to  carry  a  shear  of  2  per  cent  of  the  direct  stress  on  the  column. 

The  inclination  of  the  lattice  bars  with  the  axis  of  the  member  shall  not 
ordinarily  be  under  45  degrees.  When  the  distance  between  the  rivet  lines 
in  the  flange  is  more  than  15  inches,  if  a  single  riveted  bar  is  used  the  lattic- 
ing shall  be  double,  and  the  bars  riveted  at  their  intersection. 

Lattice  bars  with  two  rivets  shall  generally  be  used  in  flanges  wider  than 
5  inches. 

129.  Pitch  of  Lattice  Connections  along  the  Member.  —  The  ratio  of 
the  pitch  of  lattice  connections  along  a  flange  to  the  least  radius  of  gyration 
of  that  side  of  the  member  shall  not  exceed  the  ratio  of  length  to  least 
radius  of  gyration  for  the  whole  column. 

130.  Joints.  —  In  general  all  joints  in  riveted  work,  whether  for  tension 
or  compression  members,  shall  be  fully  spliced.     Joints  in  compression 
members  when  the  abutting  faces  are  finished  for  bearing  may  be  spliced 
sufficiently  to  hold  the  connecting  members  accurately  in  place. 

131.  Pins.  —  A  pin  shall  have  a  diameter  not  under  f  of  the  width  of 
the  widest  bar  held  by  it.     Pins  must  be  turned  true  to  size  and  straight; 
they  must  be  driven  with  pilot  nuts. 

132.  Pinholes.  —  Pinholes   shall  be  reinforced  by  plates   where   neces- 
sary.    At  least  one  plate  shall  be  as  wide  as  the  projecting  flanges  will 
allow;    where  angles  are  used,  this  plate  shall  be  on  the  same  side  as  the 
angles.     The  plates  must  contain  sufficient  rivets  that  their  portion  of  the 
pin  pressure  may  be  distributed  to  the  full  cross  section  of  the  member. 

Pins  must  be  sufficiently  long  to  insure  a  full  bearing  of  all  parts  connected 
upon  the  turned-down  body  of  the  pin.  Members  should  be  packed  on 
pins  to  produce  the  least  bending  moment  on  the  pin.  Vacant  spaces 


364  GRAPHICS   AND   STRUCTURAL   DESIGN 

along  the  pin  should  be  filled  and  the  members  held  against  lateral  move- 
ment. 

133.  Temperature  Range.  —  Expansion  and  contraction  shall  be  pro- 
vided for  corresponding  to  a  temperature  range  of  1 50  degrees  Fahrenheit. 

134.  Expansion  Rollers.  —  The  minimum  diameter  of  expansion  rollers 
shall  be  4  inches. 

135.  Anchor  Bolts.  —  Columns,  when  resisting  tensile  stresses  at  their 
bases,  shall  be  anchored  by  bolts  to  the  foundations.     Anchor  bolts  shall 
be  long  enough  to  secure  a  weight  of  masonry  at  least  one  and  one-half 
times  the  tension  in  the  anchor.     The  minimum  size  of  anchor  bolts  shall 
be  ij  inches. 

Materials.  —  See  paragraphs  i  to  18  inclusive. 
Workmanship.  —  See  paragraphs  19  to  57  inclusive. 
Inspection.  —  See  paragraphs  58  to  65  inclusive. 

SPECIFICATIONS   FOR  A  DECK-PLATE   GIRDER 
RAILWAY   BRIDGE 

136.  Material.  —  All  material  to  be  rolled  steel  as  specified  in  para- 
graphs i.  to  10  inclusive.     Cast  iron  or  steel  castings  will  be  permitted 
only  in  machinery  of  movable  bridges  and  in  special  cases  for  shoes  and 
bearings. 

137.  Plate  Girders.  —  Plate  girders  are  recommended  for  spans  frofn 
20  feet  to  100  feet. 

138.  Spacing  of  Girders.  —  Deck  plate  girders  shall  generally  be  spaced 
6  feet  6  inches. 

139.  Floor.  —  The   floor   shall   consist   of   8-inch   by  8-inch   cross  ties 
separated  6  inches.     They  shall  be  notched  to  fit  on  the  flanges  on  which 
they  should  have  a  full  and  even  bearing. 

140.  Guard  Rails.  —  Guard  timbers  6  by  8  inches  shall  be  placed  on 
each  side  of  the  track.     Their  inner  faces  shall  be  not  less  than  3  feet  3 
inches  from  the  center  of  the  track.     These  guard  timbers  shall  be  notched 
one  inch  over  each  tie  and  shall  be  fastened  to  every  third  tie  and  at  each 
splice  by  a  f-inch  bolt.     Splices  shall  be  over  floor  timbers  with  half-and- 
half  joints  of  6  inches  lap.     The  floor  and  guard  timbers  must  be  continued 
over  piers  and  abutments. 

LOADS 

141.  Dead  Load.  —  In  estimating  the  weight  of  the  structure,  for  use 
in  calculating  the  strains,  the  timber  shall  be  assumed  as  weighing  \\  pounds 
per  foot  B.  M.  (board  measure,  144  cubic  inches).     The  weight  of  the  rails, 
spikes  and  joints  shall  be  taken  at  160  pounds  per  lineal  foot  of  track. 

142.  Live  Load.  —  All  bridges  shall  be  designed  to  carry,  in  addition  to 
their  own  weight  and  that  of  the  floor,  a  moving  load  on  each  track  con- 


MATERIALS  365 

sisting  of  two  engines  coupled  at  the  head  of  a  uniformly  distributed  train 
load,  placed  so  as  to  give  the  greatest  strain  in  the  structure. 

This  loading  is  frequently  specified  Cooper's  £-40  as  a  minimum. 

NOTE.  —  The  loading  Cooper's  E-6o  is  given  on  page  7  1  .  For  the  other  Cooper's 
loadings  the  distances  between  the  wheels  remain  the  same  as  the  E-6o  loading. 
The  driving-wheel  loads  are  1000  times  the  number  designating  the  loading;  thus 
for  E-6o  the  load  on  the  drivers  is  60,000  pounds,  while  for  £-40  it  is  40,000  pounds. 
The  pilot-wheel  load  is  assumed  50  per  cent  of  the  load  on  the  drivers,  while  the 
tender-wheel  loads  are  65  per  cent  of  the  driving-wheel  loads. 

143.  Impact  Allowance.  —  The  effect  of  impact  and  vibration  shall  be 
added  to  the  above-mentioned  maximum  live-load  stresses,  and  shall  be 
determined  by  the  formula, 


30o 

where  I  =  impact  to  be  added  to  the  live-load  stresses; 
5  =  computed  live-load  stress; 
L  =  length  of  loaded  distance  which  produces  the  maximum  stress 

in  the  member. 

Impact  shall  not  be  added  to  stresses  produced  by  longitudinal,  cen- 
trifugal and  lateral  wind  forces. 

144.  Wind  and  Lateral  Loading.  —  All  bridges  shall  be  designed  for  a 
lateral  force  on  the  loaded  chord  of  200  pounds  per  lineal  foot  plus  10  per 
cent  of  the  specified  train  load  on  one  track,  and  200  pounds  per  lineal 
foot  on  the  unloaded  chord,  these  forces  being  considered  as  moving. 

NOTE.  —  This  loading  provides  for  the  wind  load  and  vibration  and  impact  due 
to  lateral  swaying  of  the  train  caused  by  unbalancing  of  the  locomotives,  etc. 

UNIT  STRESSES 

145.  Unit  Stresses  in  Structural  Steel.  —  All  parts  of  the  structure  shall 
be  so  designed  that  on  structural  and  rivet  steels  the  sums  of  the  maximum 
stresses  shall  not  exceed  the  following  unit  stresses,  excepting  as  specified 
in  paragraph  91. 


Character  of  stress. 


Allowable  unit  stress, 
Ibs.  per  sq.  in. 


Axial  tension,  net  section  

Flexural  stress,  extreme  fiber  stress,  net  section, 
Flexural  stress,  on  pins,  extreme  fibers. 


16,000 
16,000 
24,000 

Axial  compression,  gross  section.  .  . . 

5=i6,ooo— 


W) 


5  =  allowable  unit  stress; 
/  =  unsupported  length  of  member,  in  inches; 
r  =  least  radius  of  gyration  of  the  member,  in  inches. 


366 


GRAPHICS  AND   STRUCTURAL   DESIGN 


146.   Stresses  on  Rivets.  —  The  following  are  the  allowable  working 
stresses  in  pounds  per  square  inch  on  bolts  and  rivets  in  shear  and  bearing. 


Character  of  bolt  or  rivet. 

Unit  stress. 

Shear. 

Bearing. 

Hand-driven  field  rivets  and  turned  bolts  

9,000 
II,OOO 
12,000 

10,000 

l8,OOO 
22,000 
24,000 

Power-driven  field  rivets 

Shop-driven  rivets  and  pins 

Plate  girder  webs,  gross  section  .    . 

Expansion  rollers,  per  linear  inch,  where  d  =  the 
diameter  of  the  roller  in  inches  

600  xd 

147.   Pressure  on  Foundations.  —  The  pressures  on  masonry  piers  shall 
not  exceed  the  following  loads  in  pounds  per  square  inch. 


Pounds  per 
sq.  in. 


First-class  sandstone  or  limestone  masonry,  and  Portland 
cement  concrete,  1-2-4,  including  impact 

First-class  granite  masonry,  and  Portland  cement  mortar, 
including  impact 


400 
600 


See  also  paragraph  79. 

148.  Pressure  on  Soils.  —  See  paragraph  77. 

149.  Combined  Stresses.  —  See  paragraph  93. 

150.  Girder  Design.  —  The  depth  shall  preferably  be  not  less  than  Ti 
of  the  span.     When  made  shallower  the  flange  stresses  should  be  reduced 
so  that  the  deflection  will  not  exceed  that  of  a  girder  having  its  depth  A 
of  the  span.     Girders  may  be  designed  either  by  calculating  the  moments 
of  inertia  of  their  net  sections  or  by  assuming  the  flange  areas  as  concen- 
trated at  their  centers  of  gravity;  in  this  case  |  of  the  gross  web  section  may 
be  considered  as  flange  area;  when  this  is  assumed  the  web  splices  must  be 
designed  for  carrying  bending  as  well  as  shear. 

151.  Flange  Design.  — The  flanges  shall  be  designed  so  that  the  area  of 
the  cover  plates  shall  not  exceed  60  per  cent  of  the  total  flange  area.     See 
also  paragraph  118. 

152.  Compression  Flange.  —  The  area  of  the  compression  flange  must 
at  least  equal  the  area  of  the  tension  flange  at  the  same  section  of  the  girder. 
Where  the  unsupported  distance  along  the  girder  exceeds  16  times  the  width 
of  the  flange  the  fiber  stress  in  the  compression  flange  must  be  limited  by 


MATERIALS  367 

that  determined  by  the  following  formulae.     Where  the  flange  consists  of 
plates,  /  =  16,000  —  1 200  X  -  \  ;  where  the  flange  is  a  channel  section,  /  = 

16,000  —  (150  XT  J ;  here 

/  is  the  allowable  fiber  stress  in  pounds  per  square  inch; 

/  is  the  distance  between  lateral  supports  along  the  girder,  in  inches ; 

b  is  the  width  of  the  flange  in  inches. 

153.  Flange  Rivets.  —  The  rivets  securing  the  upper  flange  angles  to 
the  web  plate  must  be  calculated  to  resist  the  change  in  horizontal  shear, 
together  with  such  vertical  shear  due  to  wheel  concentrations  and  floor 
loads  as  may  be  transferred  to  the  web  through  the  flange  angles. 

When  the  rails  are  placed  directly  on  the  upper  flange  the  wheel  loads 
shall  be  assumed  as  distributed  over  30  inches;  if  ties  transfer  this  load  to 
the  girder,  the  load  may  then  be  considered  as  distributed  over  three  ties. 

154.  Web  Design.  —  The  web  shall  be  designed  for  the  total  maximum 
shear,  including  dead  load,  live  load  and  impact;  the  fiber  stress  on  the  gross 
section  shall  not  exceed  10,000  pounds  per  square  inch.     See  also  para- 
graph 100. 

I55-   Web  Splices.  —  See  paragraph  150. 

156.  Web  Stiffeners.  —  There  shall  be  web  stiffeners,  generally  in  pairs, 
over  bearings,  at  points  of  concentrated  loading,  and  at  other  points  where 
the  thickness  of  the  web  is  less  than  A  of  the  unsupported  distance  between 
flange  angles.     The  distance  between  stiffeners  shall  not  exceed  that  given 
by  the  following  formula,  with  a  maximum  limit  of  six  feet  and  not  greater 

than  the  depth  of  the  web;  d  =  —   (12,000  —  s). 

40 

d  =  the  clear  distance  between  stiffeners,  in  inches; 

/  =  the  web  thickness,  in  inches; 

5  =  the  unit  shearing  stress  in  the  web,  in  pounds  per  square  inch. 

157.  Bracing. — Lateral  bracing  shall  be  placed  as  close  as  possible  to 
the  plane  of  the  upper  chord  when  clearing  the  ties. 

Spans  exceeding  60  feet  shall  have  lateral  bracing  near  the  plane  of  both 
top  and  bottom  chords.  Cross  frames  shall  be  placed  at  the  ends  and  at 
intermediate  points  separated  not  more  than  20  feet. 

158.  Minimum  Sizes  of  Materials.  —  Excepting  for  fillers  the  minimum 
thickness  of  material  shall  be  f  inch.     Rivets  shall  not  be  under  f  inch 
diameter. 

159.  Rivet  Spacing.  —  See  paragraphs  120  to  126. 

1 60.  Expansion.  —  See  paragraphs  133  and  134. 


368  GRAPHICS  AND   STRUCTURAL  DESIGN 

All  spans  exceeding  80  feet  in  length  shall  have  hinged  bolsters  at  both 
ends,  and  at  one  end  turned  rollers,  running  between  planed  surfaces. 

Rollers  shall  not  be  under  4  inches  in  diameter.  Bridges  with  spans 
under  80  feet  shall  have  one  end  free  to  move  upon  planed  surfaces. 

161.  Camber.  —  The  camber  shall  be  rV  inch  for  every  10  feet. 

NOTE.  —  Camber  is  an  upward  curvature  given  the  bridge  and  should  theoreti- 
cally be  equal  and  opposite  to  that  caused  by  the  deflection  of  the  bridge  under  the 
load  for  which  it  was  designed.  It,  therefore,  serves  to  bring  the  nominally  hori- 
zontal members  into  actually  horizontal  positions  when  subjected  to  their  full  load. 
Under  full  load  then  the  several  members  will  be  in  the  same  relative  positions  as 
those  for  which  their  stresses  were  determined. 

SPECIFICATIONS  FOR   PORTLAND    CEMENT   CONCRETE 
AND   REINFORCED    CONCRETE 

162.  Cement.  —  Cement  shall  be  Portland,  and  shall  meet  the  require- 
ments of  the  standard  specifications  of  the  American  Society  for  Testing 
Materials. 

163.  Fine  Aggregates. — These  shall  consist  of  sand,   crushed  stone, 
or  gravel  screenings  graded  from  fine  to  coarse,  and  when  dry  passing  a 
screen  having  holes  |  inch  in  diameter;  it  shall  preferably  be  of  siliceous 
material,  clean,  coarse,  free  from  vegetable  loam  or  other  deleterious  matter, 
and  not  more  than  6  per  cent  shall  pass  a  sieve  having  100  meshes  per  linear 
inch. 

164.  Test.  —  Mortars  composed  of  one  part  Portland  cement  and  three 
parts  fine  aggregate  by  weight  when  made  into  briquettes  shall  show  a 
tensile  strength  of  at  least  70  per  cent  of  the  strength  of  1-3  mortar  of 
the  same  consistency  made  with  the  same  cement  and  standard  Ottawa 
sand. 

165.  Coarse    Aggregates.  —  These   shall   consist   of   crushed   stone   or 
gravel,  graded  in  size,  which  is  retained  on  a  screen  having  holes  |  inch 
in  diameter;  it  shall  be  clean,  hard,  durable  and  free  from  all  deleterious 
matter.    Aggregates  containing  soft,  flat  or  elongated  particles  shall  not  be 
used. 

166.  Maximum  Size  of  Coarse  Aggregate.  —  The  maximum  size  of  the 
coarse  aggregate  shall  be  such  that  it  will  not  separate  from  the  mortar  in 
laying  and  will  not  prevent  the  concrete  fully  surrounding  the  reinforce- 
ment or  rilling  all  parts  of  the  forms. 

Where  concrete  is  used  in  mass  the  maximum  size  of  the  coarse  aggregate 
may,  at  the  option  of  the  engineer,  be  such  as  to  pass  a  3 -inch  ring.  For 
reinforced  concrete,  sizes  usually  are  not  to  exceed  one  inch  in  any  direction, 
but  may  be  varied  to  suit  the  character  of  the  reinforcement. 


MATERIALS 


369 


167.  Water.  —  Water  used  in  mixing  concrete  shall  be  free  from  oil, 
acid,  alkalies,  or  vegetable  matter. 

REINFORCING  STEEL 

1 68.  Metal  Reinforcement.  —  The  metal  reinforcing  steel  shall  be  manu- 
factured from  new  billets  and  shall  meet  the  requirements  of  the  following 
specifications  and  be  free  from  rust,  scale,  or  coatings  of  any  character 
which  would  tend  to  reduce  or  destroy  the  bond. 

169.  Process  of  Manufacture.  —  See  paragraph  i. 

170.  Chemical  and  Physical  Requirements.  — 


Requirements. 

Structural  steel. 

High  carbon 
steel. 

T,i        1                       (  Basic 

o  04% 

o  08  q% 

Phosphorus,  max.  <   *    -^ 

o  06% 

o  071;% 

Sulphur,  maximum  

0.05% 

Ultimate  tensile  strength,  Ibs.  per  sq.  in..  . 

(      Desired 
}       60,000 
1,500,000 

Desired 
85,000 
1,400,000 

Elongation,  min.  per  cent  in  2  ins.,  Fig.  2.  .  . 

Ult.  ten.  str. 

22% 

Ult.  ten.  str. 

Character  of  fracture  

Silky 

Cold  bends  without  fracture.  . 

i  80°  flat 

180°  d  =  At 

171.  Yield  Point. — The  yield  point,  as  indicated  by  the  drop  of  the  beam, 
shall  be  not  less  than  60  per  cent  of  the  ultimate  strength. 

172.  Allowable  Variations.  —  See  paragraph  3. 

173.  Chemical  Analyses.  —  See  paragraph  9. 

174.  Forms  of  Specimens.  —  See  paragraph  10. 

175.  Number  of  Tests.  —  At  least  one  tensile  and  one  bending  test  shall 
be  made  from  each  melt  of  steel  as  rolled. 

176.  Modifications  in  Elongation.  —  See  paragraph  4. 

177.  Bending  Tests.  —  Bending  tests  may  be  made  by  pressure  or  blows. 
Shapes  and  bars  less  than  one  inch  thick  shall  bend  as  called  for  in  para- 
graph 170. 

178.  Bending  Tests  of  Thick  Material.  —  See  paragraph  5. 

179.  Finish.  —  See  paragraph  14. 

180.  Stamping.  —  See  paragraph  15. 

181.  Defective  Material.  —  See  paragraph  16. 

CONCRETE 

182.  Proportions.  —  The  materials  to  be  used  in  concrete  shall  be  of 
uniform  quality  and  so  proportioned  as  to  secure  as  nearly  as  possible  a 
maximum  density. 


370  GRAPHICS  AND  STRUCTURAL  DESIGN 

183.  Unit  of  Measure.  —  The  unit  of  measure  shall  be  the  barrel,  which 
shall  be  taken  to  be  3.8  cubic  feet.     Four  bags  containing  94  pounds  of 
cement  each  shall  be  considered  the  equivalent  of  one  barrel.     Fine  and 
coarse  aggregates  shall  be  measured  separately  as  loosely  thrown  into  the 
measuring  receptacle. 

184.  Relation  of  Fine  and  Coarse  Aggregate.  —  The  fine  and  coarse 
aggregate  shall  be  used  in  such  relative  proportions  as  will  insure  maximum 
density. 

185.  Relation   of   Cement   and   Aggregates.  —  For   reinforced-concrete 
construction  a  density  proportion  based  on  one  to  six  shall  be  used,  i.e., 
one  part  of  cement  to  a  total  of  six  parts  of  fine  and  coarse  aggregates 
measured  separately. 

1 86.  Mixture  for  Massive  Concrete.  —  For  massive  or  rubble  concrete 
a  density  proportion  based  on  one  to  nine  shall  be  used. 

187.  Mixing.  —  The  ingredients  of  concrete  shall  be  thoroughly  mixed 
to  the  desired  consistency,  and  the  mixing  shall  continue  until  the  cement 
is  uniformly  distributed  and  the  mass  is  uniform  in  color  and  homogeneous. 

1 88.  Measuring  Proportions.  —  Methods  of  measurement  of  the  pro- 
portions of  the  various  ingredients,  including  the  water,  shall  be  used,  which 
will  secure  separate  uniform  measurements  at  all  times. 

189.  Machine  Mixing.  —  When  the  conditions  will  permit  a  batch  mixer 
'of  a  type  which  insures  uniform  mixing  of  the  materials  throughout  the  mass 
shall  be  used. 

190.  Hand  Mixing.  —  When  it  is  necessary  to  mix  by  hand. the  mixing 
shall  be  on  a  watertight  platform  and  especial  precautions  shall  be  taken 
to  turn  the  materials  until  they  are  homogeneous  in  appearance  and  color. 

(a)  Tight  platforms  shall  be  provided  of  sufficient  size  to  accommodate 
men  and  materials  for  the  progressive  and  rapid  mixing  of  at  least  two 
batches  of  concrete  at  the  same  time.     Batches  shall  not  exceed  one  cubic 
yard  each,  and  smaller  batches  are  preferable,  based  upon  a  multiple  of 
the  number  of  sacks  of  cement  to  the  barrel. 

(b)  Spread  the  fine  aggregates  evenly  upon  the  platform,  and  the  cement 
upon  the  fine  aggregates;   mix  these  thoroughly  until  of  an  even  color. 
Add  all  the  water  necessary  to  make  a  thin  mortar  and  spread  again;   add 
the  coarse  aggregates,  which,  if  dry,  should  first  be  thoroughly  wet  down. 
Turn  the  mass  with  shovels  or  hoes  until  thoroughly  incorporated  and  all 
the  aggregates  are  thoroughly  covered  with  mortar;    this  will  probably 
require  the  mass  to  be  turned  four  times. 

(c)  Another  approved  method,  which  may  be  permitted  at  the  option  of 
the  engineer  in  charge,  is  to  spread  the  fine  aggregates,  then  the  cement, 
and  mix  dry,  then  the  coarse  aggregates;    add  water  and  mix  thoroughly 
as  above. 


MATERIALS  371 

191.  Water.  —  The  materials  shall  be  mixed  wet  enough  to  produce  a 
concrete  of  such  a  consistency  as  will  flow  into  the  forms  and  into  the  metal 
reinforcement,  and  which,  on  the  other  hand,  can  be  conveyed  from  the 
place  of  mixing  to  the  forms  without  separation  of  the  coarse  aggregate  from 
the  mortar. 

192.  Retempering.  —  Remixing  concrete  with  water  after  it  has  partially 
set  will  not  be  permitted. 

193.  Placing  Concrete.  —  (a)  After  the  addition  of  water,  it  shall  be 
handled  rapidly  from  the  place  of  mixing  to  the  place  of  final  deposit,  and 
under  no  circumstances  shall  concrete  be  used  that  has  partially  set  before 
final  placing. 

(b)  The  concrete  shall  be  deposited  in  such  a  manner  as  will  permit  the 
most  thorough  compacting,  such  as  can  be  obtained  by  working  with  a 
straight  shovel,  or  slicing  tool  kept  moving  up  and  down  until  all  the  in- 
gredients have  settled  in  their  proper  place  by  gravity  and  the  surplus  water 
has  been  forced  to  the  top. 

(c)  In  depositing  concrete  under  water,  special  care  shall  be  exercised 
to  prevent  the  cement  from  floating  away,  and  to  prevent  the  formation 
of  laitance. 

NOTE.  —  Laitance  is  a  whitish  gelatinous  substance,  of  about  the  same  composi- 
tion as  cement  but  having  little  or  no  hardening  properties.  It  is  caused  by  the 
action  of  water  on  the  surface  of  the  concrete. 

(d)  Before  placing  concrete  the  forms  shall  be  thoroughly  wetted  and  the 
space  to  be  occupied  by  the  concrete  freed  from  debris. 

(e)  When  work  is  resumed,  concrete  previously  placed  shall  be  roughened, 
thoroughly  cleaned  of  foreign  material  and  laitance,  drenched  and  slushed 
with  a  mortar  consisting  of  one  part  of  Portland  cement  and  not  more  than 
two  parts  of  fine  aggregate. 

(f)  The  faces  of  concrete  exposed  to  premature  drying  shall  be  kept  wet 
for  a  period  of  at  least  seven  days. 

194.  Freezing  Weather.  —  The  concrete  shall  not  be  mixed  or  deposited 
at  a  freezing  temperature,  unless  special  precautions,  approved  by  the  engi- 
neer, are  taken  to  avoid  the  use  of  materials  containing  frost  or  covered 
with  ice  crystals,  and  to  provide  means  to  prevent  the  concrete  from  freez- 
ing after  being  placed  in  position  and  until  it  is  thoroughly  hardened. 

195.  Rubble   Concrete.  —  Where   the  concrete  is   to  be   deposited  in 
massive  work,  clean  stones,  thoroughly  embedded  in  the  concrete  as  near 
together  as  is  possible  and  still  entirely  surrounded  by  concrete,  may  be 
used  at  the  option  of  the  engineer. 

196.  Forms.  —  Forms  shall  be  substantial  and  unyielding  and  built  so 
that  the  concrete  shall  conform  to  the  designed  dimensions  and  contours, 


372  GRAPHICS  AND   STRUCTURAL  DESIGN 

and  so  constructed  as  to  prevent  the  leakage  of  mortar.     These  forms  shall 
not  be  removed  until  authorized  by  the  engineer. 

197.  Forms  for  Important  Work.  —  For  important  work,  the  lumber 
used  for  face  work  shall  be  dressed  to  a  uniform  thickness  and  width,  and 
shall  be  sound  and  free  from  loose  knots,  secured  to  the  studding  or  up- 
rights in  horizontal  lines. 

198.  Less    Important    Work.  —  For   backings   and   other    rough   work 
undressed  lumber  may  be  used. 

199.  Round  Corners.  —  Where  corners  of  masonry  and  other  projections 
liable  to  injury  occur,  suitable  moldings  shall  be  placed  in  the  angles  of 
the  forms  to  round  or  bevel  them  off. 

200.  Re-using  Lumber.  —  Lumber  once  used  in  forms  shall  be  cleaned 
before  using  again. 

20 1.  Wetting  Forms.  —  In  dry  but  not  freezing  weather  the  forms  shall 
be  drenched  with  water  before  the  concrete  is  placed  against  them. 

DETAILS  OF  CONSTRUCTION 

202.  Splicing  Reinforcement.  —  Whenever  it  is  necessary  to  splice  the 
reinforcement  by  lapping,  the  length  of  lap  will  be  decided  by  the  engineer 
on  the  basis  of  the  safe  bond  stress  in  the  reinforcement  at  the  point  of 
splice.     Splices  shall  not  be  made  at  the  points  of  maximum  stress. 

203.  Joints  in  Concrete.  —  Concrete  structures,  wherever  possible,  shall 
be  cast  in  one  operation,  but  when  this  is  not  possible  the  work  shall  be 
stopped,  so  that  the  resulting  joint  shall  have  the  least  effect  on  the  strength 
of  the  structure. 

204.  Placing  Girders  and  Slabs.  —  Girders  or  slabs  shall  not  be  placed 
over  freshly  formed  walls  or  columns  without  permitting  a  period  of  at  least 
two  hours  to  elapse  to  provide  for  settlement  or  shrinkage  in  the  supports. 
Before  resuming  work  the  top  of  the  supports  should  be  thoroughly  cleansed 
of  foreign  matter  and  laitance. 

205.  Temperature  Changes.  —  In  massive  work,  such  as  retaining  walls, 
abutments,  etc.,  built   without   reinforcement,   joints   shall   be   provided 
approximately  every  50  feet  throughout  the  length  of  the  structure  to  care 
for  the  temperature  changes.     To  provide  against  the  structures  being 
thrown  out  of  line  by  unequal  settlement,  each  section  of  the  wall  may 
be  tongued  and  grooved  into  the  adjoining  section.     To  provide  against 
unsightly  cracks,  due  to  unequal  settlement,  a  joint  shall  be  made  at  sharp 
angles. 

206.  Surface  Finish.  —  The  desired  finish  of  the  surface  shall  be  deter- 
mined by  the  engineer  before  the  concrete  is  placed,  and  the  work  shall  be 
so  conducted  as  to  make  it  possible  to  secure  the  finish  desired. 


MATERIALS  .  373 

Plastering  of  surfaces  will  not  be  permitted.  • 

NOTE.  —  The  preceding  portion  of  this  specification  is  taken  from  those  of  the 

American  Railway  Engineering  and  Maintenance  of  Way  Association,  Proceedings 

of  1909. 

207.  Fireproofing.  —  For  ordinary  conditions,  it  is  recommended  that 
the  metal  in  columns  and  girders  be  protected  by  a  minimum  of  two  inches 
of  concrete;    that  the  metal  in  beams  be  protected  by  one  and  one-half 
inches  of  concrete  and  that  the  metal  in  slabs  be  protected  by  at  least  one 
inch  of  concrete.     In  monolithic  columns  a  section  around  the  column 
having  a  width  of  at  least  one  and  one-half  inches  should  be  allowed  for  fire- 
proofing  and  not  estimated  as  carrying  any  load.     All  corners  of  columns 
should  be  rounded,  as  rounded  corners  are  affected  less  by  fire  and  are  also 
less  liable  to  other  injury. 

LOADS 

208.  Vertical  Loads.  —  The  same  vertical  loads  as  those  given  under 
"  Steel  Mill  Building  Specifications  "  may  be  followed;    see  paragraphs 
68,  69,  70  and  74.     Buildings  in  cities  should  be  designed  in  accordance  with 
the  local  building  laws.     Where  such  laws  are  not  available  the  following 
floor  loads  may  be  assumed: 

Office  floors 75  Ibs.  per  sq.  ft. 

Floors  for  light  running  machinery 75-150  Ibs.  per  sq.  ft. 

Floors  for  medium  heavy  running 

machinery 200  Ibs.  per  sq.  ft. 

Storage,  to  be  estimated  from  weight  of 

the  materials  in  each  case 150-1000  Ibs.  per  sq.  ft. 

The  weight  of  the  concrete  shall  be  assumed  as  150  pounds  per  cubic  foot. 
For  Loads  on  Bridges  see  paragraphs  141,  142  and  143. 

209.  Wind  Loads.  —  These  will  be  the  same  as  those  given  under  the 
previous  specifications;  see  paragraphs  71,  72,  73  and  144. 

210.  Impact.  —  Impact  may  be  taken  into  account  by  making  additions 
to  the  live-load  stresses. 

211.  Dimensions  for  Calculations.  —  (a)  The  spans  of  beams  and  girders 
shall  be  assumed  as  the  distance  center  to  center  of  supports  but  need  not 
be  taken  greater  than  the  clear  span  plus  the  depth  of  the  beam  or  girder. 

(b)  For  slabs  supported  at  the  ends  the  span  shall  be  assumed  as  the 
clear  span  plus  the  depth  of  the  slab. 

(c)  For  continuous  slabs  the  spans  shall  be  taken  as  the  distances  between 
the  centers  of  supports. 

212.  Bending   Moments.  —  (a)   For   continuous  beams   and  slabs   the 
bending  moments  at  the  center  and  at  the  supports  shall  be  assumed  as 


374 


GRAPHICS   AND   STRUCTURAL  DESIGN 


80  per  cent  of  the  moment  on  a  freely  supported  beam  having  the  same  span 
and  load. 

NOTE.  —  The  Specifications  of  the  Joint  Committee  of  the  A.  S.  C.  E.,  A.  S.  T.  M., 
A.  R.  E.  and  M.  of  W.  Assn.,  etc.,  give  this  as  67  per  cent  for  interior  spans  but 
80  per  cent  for  end  spans.  In  the  case  of  beams  and  slabs  continuous  for  two 
spans  only,  the  bending  moment  at  central  support  shall  be  calculated,  using  100 
per  cent  as  the  factor,  while  near  the  middle  of  the  span  the  factor  shall  be  assumed 
80  per  cent. 

(b)  Floor  slabs  should  be  designed  and  reinforced  as  continuous  over 
the  supports.  If  the  length  of  the  slab  exceeds  one  and  one-half  times  its 
width,  the  entire  load  shall  be  carried  by  the  transverse  reinforcement. 

Square  slabs  shall  be  reinforced  in  both  directions. 

NOTE.  —  The  distribution  of  loads  in  the  two  directions  upon  slabs  having  a 
ratio  of  length  to  width  not  exceeding  i  to  1.5  will  be  approximately  as  given  in 
the  following  table. 


I 

b' 

r. 

/ 
b' 

r. 

I  .O 

0.50 

i-3 

0-75 

I  .  I 

0-59 

i-4 

0.8o 

1.2 

0.67 

i-5 

0.83 

Here 

r  is  the  proportion  of  the  load  carried  by  the  transverse  reinforcement, 

/  is  the  length,  and 

b  is  the  breadth  of  the  slab. 

Using  the  values  just  specified,  each  set  of  reinforcements  is  to  be  calculated 
in  the  same  manner  as  for  slabs  having  supports  on  two  sides  only,  but  the 
total  amount  of  reinforcement  thus  determined  may  be  reduced  25  per  cent 
by  gradually  increasing  the  rod  spacing  from  the  third  point  to  the  edge 
of  the  slab. 

213.  T  Beams.  —  In  beam-and-slab  construction,  an  effective  bond 
should  be  provided  at  the  junction  of  the  beam  and  slab.  When  the  prin- 
cipal slab  reinforcement  is  parallel  to  the  beam,  transverse  reinforcement 
should  be  used,  extending  over  the  beam  and  well  into  the  slab. 

Where  adequate  bond  and  shearing  resistance  between  the  slab  and  web 
of  beam  is  provided,  the  slab  may  be  considered  as  an  integral  part  of  the 
beam,  but  its  effective  width  shall  be  determined  by  the  following  rules: 
(a)  It  shall  not  exceed  one-fourth  of  the  span  length  of  the  beam;  (b)  its 
overhanging  width  on  either  side  of  the  web  shall  not  exceed  four  times  the 
thickness  of  the  slab. 


MATERIALS  375 

In  the  design  of  T  beams  acting  as  continuous  beams,  careful  considera- 
tion must  be  given  to  the  compressive  stresses  at  the  supports. 

Beams  in  which  the  T  form  is  used  only  for  the  purpose  of  providing 
additional  compression  area  of  concrete  should  preferably  have  a  width  of 
flange  not  more  than  three  times  the  width  of  the  stem  and  a  thickness  of 
the  flange  not  less  than  one-third  of  the  depth  of  the  beam. 

Both  in  this  form  and  in  the  beam  and  slab  form,  the  web  stresses  and 
the  limitations  in  placing  the  longitudinal  reinforcements  will  probably  be 
controlling  factors  in  the  design. 

214.  Internal    Stresses.  —  The   internal   stresses   in  slabs,   beams   and 
girders  shall  be  determined  by  the  formulae  recommended  by  the  Joint 
Committee  of  the  A.  S.  C.  E.,  etc.,  which  is  as  given  in  the  body  of  the  book, 
page  209.     Throughout  the  entire  beam  the  shearing  and  bonding  stresses 
must  be  determined  and  proper  provision  made  to  develop  the  required 
strength. 

215.  Columns.  —  The   ratio   of   the   unsupported   length  of  a   column 
divided  by  its  least  width  shall  not  exceed  15.     The  effective  area  of  a 
column  shall  be  considered  as  the  area  within  the  fireproofing,  or  in  the  case 
of  hooped  columns,  or  columns  reinforced  with  structural  shapes,  it  shall 
be  taken  as  the  area  within  the  hooping  or  structural  shapes. 

The  following  working  fiber  stresses  may  be  allowed  in  columns  varying 
with  the  class  of  reinforcement  in  it. 

(a)  Columns  with  longitudinal  reinforcement  only,  such  reinforcement 
not  less  than  i  per  cent  nor  more  than  4  per  cent.    The  unit  stress  shall  be 
those  allowed  for  axial  compression  in  Working  Stresses,  paragraph  218. 

(b)  Columns  with  reinforcements  of  bands,  hoops  or  spirals,  as  hereafter 
specified :  the  stresses  allowed  shall  be  20  per  cent  higher  than  those  given 
for  (a),  provided  the  ratio  of  the  unsupported  length  of  the  column  to 
the  diameter  of  the  hooped  core  is  not  more  than  eight. 

(c)  Columns  reinforced  with  not  less  than  i  per  cent  and  not  more  than 
4  per  cent  of  longitudinal   bars,  and  with  hoops  or  spirals,  as  hereafter 
specified:   the  stresses  shall  be  45  per  cent  higher  than  those  given  for  (a), 
provided  the  ratio  of  the  unsupported  length  of  the  column  to  the  diameter 
of  the  hooped  core  is  not  more  than  eight. 

NOTE.  —  In  all  cases  the  longitudinal  reinforcement  is  supposed  to  carry  its  pro- 
portion of  the  stress.  The  hoops  or  bands  are  not  to  be  counted  on  as  adding 
directly  to  the  strength  of  the  column. 

Bars  composing  longitudinal  reinforcement  shall  be  straight  and  shall  have  suffi- 
cient lateral  support  to  be  securely  held  in  place  until  the  concrete  has  set. 

Where  hooping  is  used,  the  total  amount  of  such  reinforcement  shall  not  be  less 
than  i  per  cent  of  the  volume  of  the  column  inclosed.  The  clear  spacing  of  such 
hooping  shall  not  be  greater  than  one-sixth-  of  the  diameter  of  the  inclosed  column, 


376 


GRAPHICS   AND   STRUCTURAL  DESIGN 


and  preferably  not  greater  than  one-tenth;  in  no  case  shall  it  exceed  2.5  inches. 
The  hooping  must  be  circular,  and  the  ends  of  bands  must  be  united  to  develop 
the  full  strength  of  the  bands.  Adequate  means  must  be  provided  to  hold  the 
hoops  or  bands  in  place  so  that  the  column  shall  have  a  straight  and  well-centered 
core.  As  the  effect  of  hooping  decreases  rapidly  with  the  increase  of  ratio  of  length 
to  core  diameter  the  above  fiber  stresses  are  limited  to  columns  in  which  this  ratio 
does  not  exceed  8. 

216.  Bending  Stresses  on  Columns.  —  Bending  stresses  due  to  eccentric 
loading  and  lateral  forces  must  be  provided  for  by  increasing  the  section 
until  the  maximum  stress  does  not  exceed  the  values  above  specified;   and 
where  tension  is  possible  in  the  longitudinal  bars,  adequate  connection 
between  the  ends  of  the  bars  must  be  provided  to  take  this  tension. 

217.  Reinforcing   for    Shrinkage    and   Temperature    Stresses.  —  Rein- 
forcement not  under  one-third  of  i  per  cent,  and  able  to  develop  a  high 
bonding  strength  shall  be  provided;    it  shall  be  placed  near  the  exposed 
surfaces  and  be  well  distributed. 


WORKING  STRESSES 

218.  Working  Fiber  Stresses.  —  The  following  working  stresses  for 
concrete  are  based  on  the  compressive  strength  developed  by  the  concrete 
after  28  days,  when  tested  in  cylinders  8  inches  in  diameter  and  16  inches 
long.  Such  tests  should  show  the  following  ultimate  strengths,  in  pounds 
per  square  inch. 


Aggregate. 

1:1:2 

i  :  1.5:3 

1:2:4 

i:2.5:5 

i:3:6 

Granite,  trap  rock  

3300 

2800 

22OO 

I8OO 

I4.OO 

Gravel,  hard  limestone  and  hard 
sandstone  

3000 

2<OO 

2OOO 

I6OO 

I  3OO 

Soft  limestone  and  sandstone  
Cinders 

2  2OO 

800 

I800 
7OO 

I50O 
600 

I2OO 
COO 

IOOO 
4OO 

(a)  Bearing.  —  When  compression  is  applied  to  a  surface  of  concrete  of 
at  least  twice  the  loaded  area,  a  stress  of  32.5  per  cent  of  the  ultimate  com- 
pressive strength  may  be  allowed. 

(b)  Axial  Compression.  —  For  concentric  compression  on  a  plain  con- 
crete column  or  pier,  the  length  of  which  does  not  exceed  12  diameters, 
22.5  per  cent  of  the  ultimate  compressive  strength  of  the  concrete  may  be 
allowed.     See  also  paragraph  215. 

(c)  Compression  in  Extreme  Fibers.  —  The  extreme  fiber  stress  on  a 
beam,  calculated  on  the  assumption  of  a  constant  modulus  of  elasticity  for 
the  concrete  under  working  stresses,  may  reach  32.5  per  cent  of  the  ultimate 


MATERIALS 


377 


compressive  strength  of  the  concrete.     Adjacent  to  the  supports  of  con- 
tinuous beams,  these  stresses  may  be  increased  15  per  cent. 

(d)  Shear  and  Diagonal  Tension.  —  In  calculations  on  beams  in  which 
the  maximum  shearing  stress  in  a  section  is  used  as  the  means  of  measur- 
ing the  resistance  to  diagonal  tension  stress,  the  following  allowable  values 
for  the  maximum  vertical  shearing  stresses  are  recommended. 

1.  For  beams  with  horizontal  bars  only  and  without  web  reinforcement, 
2  per  cent  of  the  ultimate  compressive  strength  may  be  allowed. 

2.  For  beams  thoroughly  reinforced  with  web  reinforcement,  the  value 
of  the  shearing  stress  having  been  calculated,  using  the  total  vertical  shear 
in  determining  the  unit  horizontal  shear,  the  working  shearing  stress  should 
not  exceed  6  per  cent  of  the  ultimate  compressive  strength  of  the  concrete. 
The  web  reinforcements,  exclusive  of  the  bent-up  bars  in  this  case,  shall 
be  proportioned  to  resist  two-thirds  of  the  external  vertical  shear. 

3.  For  beams  in  which  part  of  the  longitudinal  reinforcement  is  used 
in  the  form  of  bent-up  bars  distributed  over  a  portion  of  the  beam  in  a  way 
covering  the  requirements  of  this  type  of  web  reinforcement:    the  limit  of 
the  allowable  working  shearing  stress  shall  be  3  per  cent  of  the  ultimate 
compressive  strength  of  the  concrete. 

4.  Where  punching  shear  occurs,  that  is,  shearing  stress  uncombined 
with  compression  normal  to  the  shearing  surface,  and  with  all  tension  normal 
to  the  shearing  plane  provided  for  by  reinforcement;    a  shearing  stress  of 
6  per  cent  of  the  ultimate  compressive  strength  of  the  concrete  may  be 
allowed. 

(e)  Bond.  —  The  bond  stress  between  plain  reinforcing  bars  and  concrete 
may  be  assumed  as  4  per  cent  of  the  ultimate  compressive  strength  of  the 
concrete.     In  the  case  of  drawn  wire,  2  per  cent  should  be  the  limit. 

219.  Steel.  —  The  working  fiber  stress  in  the  steel  shall  not  exceed 
16,000  pounds  per  square  inch. 

220.  Modulus  of  Elasticity.  —  The  ratio  of  the  moduli  of  elasticity  of 
steel  and  concrete  will  vary  with  the  ultimate  compressive  strength  of  the 
concrete.     The  following  values  may  be  used. 


Strength  of  concrete,  Ibs.  per  sq.  in. 

Ratio  of  moduli 
of  elasticity, 
concrete  to 
steel. 

2  200  and  less  
Exceeding  2200  but  under  2900  

rV 

Exceeding  2900 

i 

CHAPTER   XXII 


PROBLEMS 

THE  object  of  these  problems  is  to  furnish  work  for  the  student  paralleling 
that  in  the  drawing  room.  Introductory  review  problems  in  Mechanics 
are  given  that  lead  up  to  the  principal  problems  in  the  book.  As  far  as  pos- 
sible, all  the  problems  have  been  made  similar  to  those  occurring  frequently 
in  practice.  Although  the  tables  of  sections  in  the  book  give  sufficient  data 
for  the  working  of  the  problems  the  writer  believes  that  the  small  additional 
outlay  for  one  of  the  handbooks  issued  by  the  manufacturers  of  structural 
steel  is  money  well  spent.  Among  those  issuing  these  books  are  the  Cambria 
Steel  Company  of  Johnstown,  Pa.,  and  the  Carnegie  Steel  Company  and 
Jones  &  Laughlin  Company  of  Pittsburgh,  Pa. 

CENTER  OF  GRAVITY 

1.  Prove  that  the  distance  from  the  base  to  the  center  of  gravity  of  a 
rectangle  is  £  h. 

2.  Prove  that  the  distance  from  the  base  to  the  center  of  gravity  of  a 
triangle  is  \  h. 

3.  Prove  that  the  distance  from  the  base  to  the  center  of  gravity  of  a 
half-round  section  is  c  =  0.4244  R. 

NOTE.  —  Use  the  fact  established  in  Problem  i  and  make  the  infinitesimal 
strips  run  from  the  diameter  to  the  circumference. 


b— * 


IK" 

f 

f 

j 

i 
8" 
i 

1 

«-3' 

| 

• — 8" H 

FIG.  i.  FIG.  2. 

4.   Prove  that  in  the  given  trapezoid,  Fig.  i, 


FIG.  3. 


PROBLEMS 


379 


5.  Calculate  the  distance  from  the  back  to  the  center  of  gravity  of  a 
i5-in.  channel  weighing  33  Ibs.  per  ft. 

6.  In  a  6  X  4  X  |-in.  angle  calculate  the  distance  of  the  center  of  gravity 
from  the  back  of  the  short  leg. 

7.  How  far  from  the  base  is  the  center  of  gravity  in  Fig.  2? 

8.  How  far  from  the  base  is  the  center  of  gravity  in  Fig.  3? 

o  f 

9.  Prove  that   the  center  of  gravity  of  a  semicircular  arc 

from  the  diameter  joining  its  extremities. 

10.  How  far  from  its  base  is  the  center  of  gravity  of  a  cone? 
from  its  base  is  the  center  of  gravity  of  its  lateral  surface? 

MOMENTS  OF  INERTIA 


is  c  =  — 

TT 


How  far 


••  f'fdA. 


Given  /  the  moment  of  inertia,  then  /'  referred  to  any  axis  parallel  to 
the  principal  axis  is  /'  =  /  +  AW. 

In  these  problems  do  not  recalculate  positions  of  centers  of  gravity. 

ii.  Calculate  the  moment  of  inertia  about  the  principal  axis  of  a  rec- 
tangle of  altitude  h  and  width  6. 


1— 


2 
FIG.  4. 


FIG.  5. 


FIG.  6. 


12.  Calculate  the  moment  of  inertia  of  a  triangle  about  an  axis  through 
its  axis  center  of  gravity  and  parallel  to  its  base. 

13.  Calculate  the  moment  of  inertia  of  a  circle  for  an  axis  through  its 
center.     (Use  polar  coordinates.) 

14.  Calculate  the  moment  of  inertia  of  a  6  X  4  X  f -in.  angle  referred  to 
an  axis  through  its  center  of  gravity  and  parallel  to  its  short  leg. 

15.  Calculate  the  inertia  of  a  i2-in.  channel  at  20^  Ibs.  per  ft.  for  an  axis 
through  its  center  of  gravity  and  parallel  to  its  back. 

16.  Calculate  the  inertia  of  two  lo-in.  channels  at  1 5  Ibs.  per  ft.  referred 
to  axis  2-2  when  the  distance  between  channel  backs  is  6.33  ins.    (See  Fig.  4.) 

17.  Calculate  the  inertia  about  axis  i-i  for  two  5  X  3  X  Hn.  angles 
placed  as  in  Fig.  5. 


38o 


GRAPHICS   AND   STRUCTURAL   DESIGN 


18.  Calculate  the  inertia  for  axis  i-i  passing  through  center  of  gravity  of 
Fig.  2. 

19.  Calculate  the  inertia  for  axis  i-i  of  four  4  X  4  X  £-in.  angles  placed 
as  in  Fig.  6. 

20.  Calculate  the  inertia  of  a  half-round  for  axis  i-i,  Fig.  7. 


k— r- 


TT 
i_ 


T 


L 


FIG.  7. 


FIG.  9. 


21.  Calculate  the  inertia  of  the  girder  section  referred  to  axis  i-i,  Fig.  8. 
Web  |-in.  plate. 

Angles  4  X  4  X  f  in. 
Flange  plates  10  X  TV  in. 

Assume  the  section  contains  two  yt"m-  diam.  holes  for  f-in.  rivets  through 
horizontal  legs  of  angles  and  flange  plates  for  which  allowance  must  be  made. 

REACTIONS  AND  BENDING  MOMENTS 

It  should  be  recalled  that  for  equilibrium  when  the  forces  act  on  a  section 
or  a  point  that  the 

Sum  of  the  horizontal  forces  =  o. 
Sum  of  the  vertical  forces  =  o. 
Sum  of  the  moments  of  forces  =  o. 

The  vertical  shear  at  a  section  is  the  algebraic  sum  of  the  vertical  forces 
to  the  left  of  that  section. 

22.  A  cantilever  beam  of  length  Z,  ft.  carries  the  load  P  at  its  free  end. 
What  is  the  reaction  at  the  support,  bending  moment  at  this  point  and  also 
bending  moment  a  distance  X  from  free  end? 

23.  A  cantilever  beam  of  length  L  ft.  carries  a  uniform  load  of  W  Ibs. 
per  ft.     What  is  the  reaction,  the  bending  moment  at  the  support  and  also 
at  a  distance  X  ft.  from  the  free  end? 

24.  A  cantilever  beam  carries  a  load  of  W  Ibs.,  varying  uniformly  from 
zero  at  the  free  end  to  a  maximum  at  the  support.     Required  the  reaction, 
the  bending  moment  at  the  support  and  at  a  distance  X  ft.  from  the  free  end. 

25.  A  simple  beam  of  span  Z,  ft.  carries  a  central  load  of  W  Ibs.     What 
are  the  reactions,  the  maximum  bending  moment  and  the  bending  moment 
X  ft.  to  the  right  of  the  left  support? 


PROBLEMS 


26.  A  simple  beam  carries  a  uniform  load  of  W  Ibs.  per  ft.     Its  span  is  L  ft. 
What  are  the  reactions,  the  maximum  bending  moment  and  the  bending 
moment  X  ft.  to  the  right  of  the  left  support? 

27.  A  simple  beam,  Fig.  9,  carries  a  load  W,  varying  from  zero  at  sup- 
ports to  a  maximum  at  the  middle.    Required  the  reactions,  bending  moment 
at  the  middle  and  at  a  distance  A'  ft.  from  the  left  support. 

28.  A  simple  beam,  Fig.  10,  carries  a  load  W  Ibs.,  varying  from  zero 
at  the  left  support  to  a  maximum  at  the  right  support.     Required  the  reac- 
tions, the  maximum  bending  moment  and  the  distance  it  occurs  from  the 
left  support. 


k-io 


FIG.  10. 


FIG.  ii. 


-30-  • 

FIG.  12. 


29.  Required  the  reactions,  maximum  bending  moment,    and   bending 
moment  X  ft.  from  the  left  support  in  Fig.  n. 

30.  In  Fig.  12  find  the  reactions  and  maximum  bending  moment.     What 
is  the  bending  moment  15  ft.  from  the  left  support? 

31.  In  Fig.  13  what  are  the  reactions?     What  is  the  bending  moment 
over  support?     What  is  the  bending  moment  X  ft.  to  the  right  of  the  left 
support?     What  is  the  maximum  bending  moment  and  where  does  it  occur? 


FIG.  13. 


FIG.  14. 


FIG.  15. 


32.  The  given  beam  in  Fig.  14  in  addition  to  the  loads  shown  carries 
a  uniform  load  of  500  Ibs.  per  ft.  of  span.     What  is  the  maximum  bending 
moment  and  where  does  it  occur?     What  is  the  bending  moment  28  ft. 
from  the  left  support? 

33.  Prove  that  in  Fig.  15  when  two  loads  P,  a  constant  distance  a  apart, 
roll  across  a  girder  the  maximum  bending  occurs  when  a  load  is  a  distance 

-  from  the  center  of  the  span. 


Since  M  =  •*—  >     it  follows    — -  =  - ; 


382  GRAPHICS  AND   STRUCTURAL  DESIGN 

SELECTION  or  BEAM  SECTIONS 

In  selecting  beams  from  manufacturers'  handbooks  it  is  common  practice 
to  use  a  factor  called  the  section  modulus  which  is  tabulated  in  the  hand- 
books. The  section  modulus  =  -or  moment  of  inertia  divided  by  the 

e 

distance  from  the  neutral  axis  to  the  extreme  fibers. 

M_ 

f 

M 
hence,  first  find  —  by  dividing  the  maximum  bending  moment    in   inch 

pounds  by  the  allowable  working  fiber  stress.  In  the  following  problems 
assume  the  beams  sufficiently  stiffened  laterally.  Problems  where  the  ratio 
of  span  to  flange  width  must  be  considered  will  be  taken  up  later. 

34.  Select  a  beam  to  carry  a  uniform  load  of  1500  Ibs.  per  lineal  foot  on  a 
span  of  20  ft.;  allow  a  working  fiber  stress  of  15,000  Ibs.  per  sq.  in. 

35.  A  beam  of  32-ft.  span  carries  four  loads  of  5000  Ibs.  each,  spaced  8  ft. 
apart,  while  the  first  is  4  ft.  from  the  left  support.     Allow  15,000  Ibs.  fiber 
stress  per  sq.  in.  and  select  an  economical  beam  section,  neglecting  the 
bending  due  to  the  beam's  weight. 


FIG.  16. 


36.  In  Fig.  1 6  assume  also  a  uniform  load  including  weight  of  beam  as 
loo  Ibs.  per  lineal  foot.     Select  an  economical  beam  section,  allowing  a 
working  fiber  stress  of  12,000  Ibs.  per  sq.  in. 

37.  In  Fig.  17  neglect  weight  of  beam,  allow  12,000  Ibs.  fiber  stress  and 
select  a  suitable  section. 

38.  The  loading  is  as  shown  in  Fig.  18.     Neglect  weight  of  beam.    Allow 
a  working  fiber  stress  of  12,000  Ibs.  per  sq.  in.  and  select  an  economical 
beam  section. 

39.  Select  I  beam  for  Fig.  19.     Assume  beam  secured  laterally.     The 
moving  wheel  loads  are  6  ft.  o  ins.  c-c.     Each  wheel  load  is  10,000  Ibs. 
Use  a  working  fiber  stress  of  12,000  Ibs.  per  sq.  in.    Assume  dead  load 
including  weight  of  beam  as  60  Ibs.  per  ft. 

40.  Select  I  beam  for  the  span  in  Fig.  20.    Assume  beam  secured  later- 
ally.    The  moving  wheel  loads  are  12  ft.  o  ins.  c-c.     Each  wheel  load  is 
1 2 ,000  Ibs.     Allowable  working  fiber  stress  is  1 2 ,000  Ibs.  per  sq.  in.     Assume 
dead  load  including  weight  of  beam  as  80  Ibs.  per  ft. 


PROBLEMS 


DEFLECTION  OF  BEAMS 


383 


It  is  not  proposed  to  review  the  entire  discussion  of  deflection  of  beams 
but  simply  to  recall  the  subject  by  one  or  two  of  the  easier  cases  and  then 
solve  a  few  problems  by  the  use  of  the  formulae  given  on  page  7.  The 
general  equation  of  the  elastic  curve  for  all  beams  is 


dx* 


El 


here  M  =  bending  moment  due  to  the  external  forces  at  a  section  whose 

abscissa  is  X. 

I  =  moment  of  inertia  of  beam  section. 
E  —  modulus  of  elasticity  of  the  material. 
y  =  ordinate  along  which  deflection  is  measured. 


41.  A  cantilever  beam,  Fig.  21,  of  uniform  section  carries  a  concentrated 
load  W  at  its  free  end.  What  is  its  deflection  if  its  length  is  L? 

Assume  origin  of  coordinates  at  o.  Here  M  =  WX  and  the  general 
equation  becomes 


ax2 
Integrate  this  twice  and  eliminate  the  constants  of  integration  by  means 

of  the  following  facts.     At  the  fixed  end,  -^  =  o  and  X  =  L,  also  Y  =  o 

ax 

when  X  =  o. 
42.   Prove  that  the  deflection  of  a  simple  beam  of  span  L  and  central 

load  W  is 

Wl? 


A  = 


48  El 


Find  the  form  of  the  equation  of  the  elastic  curve  for  this  beam,  Fig.  22, 
and  loading.     Integrate  twice,  remembering  that 

-*  =  o  when  X  =  -  and  that  X  =  o  when  y  —  o. 
dx  2 

43.   A  standard  i5-in.  I  beam  42  Ibs.  per  ft.  having  a  span  of  30  ft.  is  to 
carry  a  uniform  load;  its  working  fiber  stress  is  not  to  exceed  16,000  Ibs. 


384  GRAPHICS  AND   STRUCTURAL  DESIGN 

per  sq.  in.  and  the  maximum  deflection of  its  span.     What  load  will  it 

360 

carry? 

44.  A  6-in.  I  beam  12^  Ibs.  per  ft.  extends  5  ft.  from  a  wall  as  a  cantilever 
beam.     Assume  that  it  is  amply  stiff,  laterally.     What  load  will  it  carry 
if  the  working  fiber  stress  is  12,000  Ibs.  per  sq.  in.  and  what  will  be  its 
maximum  deflection? 

45.  What  uniform  load  will  a  i2-in.  I  beam  31  \  Ibs.  per  ft.  carry  on  a 
2o-ft.  span  if  it  is  merely  supported  at  the  ends  and  the  fiber  stress  is  16,000 
Ibs.  per  sq.  in.?      Compare  the  deflection  and  fiber  stress  of  this  beam 
with  a  beam  having  fixed  ends,  carrying  the  same  uniform  load. 

NOTE.  —  In  practice  one  generally  assumes  simple  beams,  beams  merely 
supported  at  the  reactions.  Fixed  and  continuous  beams  are  seldom  de- 
signed. Reinforced  concrete  designers  make  allowance  for  fixing  and  con- 
tinuity in  their  beam  designs. 

TENSION  PIECES 

In  the  usual  tension  piece  of  constant  cross  section  the  stress  is  uniformly 
distributed  over  the  cross  section  and  is  the  same  for  all  sections.  The 
section  may  not  be  constant,  in  which  case  the  minimum  or  net  section  must 
be  considered. 

46.  A  2-in.  round  bar  is  to  have  its  end  upset  for  U.  S.  Standard  screw 
thread.     What  size  screw  must  be  cut  if  the  area  at  the  root  of  the  thread 
exceeds  the  area  at  the  body  of  the  rod  by  18  per  cent? 

47.  A  4  X  4  X  f-in.  angle  carries  a  load  of  35,000  Ibs.  in  tension.     Assume 
two  if-in.  diameter  holes  in  a  cross  section.     What  is  the  unit  stress  per 
square  inch  on  the  net  section? 

COLUMNS 

In  the  case  of  columns  the  stress  tends  to  increase  any  initial  flexure  due 
to  inaccurate  workmanship  or  vibration  and  a  column,  when  of  sufficient 
length,  fails  by  bending.  It  should  be  noted  that  tension  in  a  piece  reduces 
any  buckling  tendency. 

COLUMNS  FAILING  BY  FLEXURE 

The  derivation  of  column  formulae  is  based  upon  the  general  equation 
of  the  elastic  curve  previously  used  in  finding  the  deflection  of  beams.  In 
these  column  formulae,  of  which  there  are  several,  a  quantity  depending 

entirely  upon  the  section  of  the  column  occurs;  it  is—  ,  and  its  square  root 

A 


PROBLEMS  385 

is  called  the  radius  of  gyration  of  the  section  referred  to  the  same  axis  as  7. 
Here  I  is  the  moment  of  inertia  and  A  the  area  of  the  section. 

_,  4  /Inertia  of  section      4 // 

Radius  of  gyration  =  v — : =  V  —  • 

*    Area  of  section         *  A 

48.  Two  8-in.  channels  1 1  i  Ibs.  per  ft.  are  spaced  such  a  distance  back 
to  back  that  the  moments  of  inertia  referred  to  their  two  principal  axes 
are  equal.     What  do  the  radii  of  gyration  equal? 

49.  Two  6  X  4  X  ^-in.  angles  are  placed  with  their  long  legs  parallel 
and  separated  f  in.     What  are  the  radii  of  gyration  referred  to  the  principal 
axes? 

50.  Two  5  X  3s  X  i-in.  angles  are  placed  with  their  short  legs  parallel 
and  separated  f  in.     What  are  the  radii  of  gyration  referred  to  the  two 
principal  axes? 

51.  Calculate  the  radius  of  gyration  of  a  circular  section  8  ins.  in  diameter. 
Then  calculate  the  radius  of  gyration  of  a  ring  section  8  ins.  outside  diam- 
eter and  6  ins.  inside  diameter. 

52.  Prove  the  following  values  of  radii  of  gyration: 

Rectangle,  principal  axis  r  = 


Triangle,  principal  axis     r  =  -  —  . 

Vi8 

Circle,  principal  axis         r  =  -• 

4 

The  following  formulae  are  frequently  used  for  mild-steel  column  design: 

Ends.  Fixed.  Hinged. 

-r,       ,  .       ,      ,,  l6,000  ,,  l6,000 

Rankme's  /  = —- ,       /  = • — . 

I"  I 


40,000  r2  10,000  r2 

Straight  line/'  =  16,000  —  35  -,      /'  =  16,000  —  70  -. 
The  most  rational  formula  for  columns  is  that  of  Ritter, 

/' A 


386  GRAPHICS  AND   STRUCTURAL  DESIGN 

/'  =  allowable  unit  working  stress,  pounds  per  square  inch. 
/  =  length  of  column,  inches. 
r  =  least  radius  of  gyration. 
/  =  maximum  desired  fiber  stress,  occurring  at  the  dangerous  section  of 

the  column  and  resulting  from  the  flexure  of  the  column. 
fe  =  fiber  stress  per  square  inch  at  the  elastic  limit  of  material. 
E  =  coefficient  of  elasticity  of  the  material  of  the  column. 
m  =  a  constant,  depending  upon  the  way  the  column  ends  are  secured. 
Its  value  for  the  usual  end  conditions  are: 

Ends.  Both  Fixed.  i  Fixed  and  i  Hinged.  Both  Hinged 

m  4  2.25  i 

For  soft  or  mild  steel,  taking  E  =  30,000,000, 

7T2  =  10  approximately 
and 

pe  =  30,000; 

the  formula  becomes,  for  hinged  ends, 

16,000 


P'  = 


10,000  \  r 


53.  Compare  the  allowable  working  stresses  as  given  by  the  three  for- 
mulae for  a  mild-steel  hinged  column,  whose  length  is  20  ft.  and  least  radius 
of  gyration  of  the  column  section  is  r  =  2. 

54.  Compare  the  allowable  working  stresses  as  given  by  the  three  for- 
mulae for  a  mild-steel  fixed-end  column  whose  length  is  20  ft.  and  least  radius 
of  gyration  of  the  column  section  is  r  =  2. 

55.  Two  i5-in.  channels  weighing  33  Ibs.  per  ft.  are  to  be  made  into  a 
latticed  column.     Assume  hinged  ends,  use  Ritter's  formula  and  determine 
the  maximum  load  which  they  can  be  designed  to  carry  if  the  column  is 
35  ft.  long. 

56.  What  load  will  a  Bethlehem  i2-in.  H  section  weighing  78  Ibs.  per  ft. 
carry?     Assume  the  column,  with  fixed  ends,  25  ft.  long  and  the  material 
mild  steel.     Use  Ritter's  formula  and  compare  the  result  with  that  given 
in  the  Bethlehem  Steel  Company's  handbook,  using  the  formula 

p'  =  16,000—  55-. 

57.  Compare  the  loads  that  can  be  carried  by  two  mild  steel  columns, 
each  22  ft.  long,  both  having  fixed  ends,  the  first  being  a  latticed  column 
made  of  two  lo-in.  channels  15  Ibs.  per  ft.,  spaced  for  maximum  load,  the 


PROBLEMS  387 

other  being  a  lo-in.  column  54  Ibs.  per  ft.,  Bethlehem  Steel  Company's 
section.     Use  Ritter's  formula  in  both  cases. 

In  structural  design  it  is  usual  practice  to  limit  the  value  of  -  for  impor- 
tant work  subjected  to  considerable  shock  to  -  =  100,  while  for  other  work 

the  usual  limit  is-  =  120. 
r 

58.  Would  a  6   X  6   X  f-in.  angle  used  singly  make  a  good  column? 
Should  it  be  used  for  a  height  of  16  ft.?     Why?     What  load  could  it  carry 
if  10  ft.  long,  with  hinged  ends?     Use  Ritter's  formula  and  assume  the 
material  mild  steel. 

59.  Would  an  8-in.  I  beam  weighing  18  Ibs.  per  ft.  make  a  good  column? 
Why?     Should  it  be  used  for  a  height  of  12  ft.?    Why?    What  load  would 
it  carry  if  8  ft.  high?     Use  Ritter's  formula.     Assume  fixed  ends  and  the 
material  mild  steel. 

60.  Two  6  X  4  X  f-in.  angles  are  to  be  placed  back  to  back,  separated 
\  in.  and  used  for  a  hinged  strut.     Take  your  data  from  Manufacturer's 
Handbook  and  show  if  it  would  be  more  economical  to  have  the  long  or 
short  legs  parallel.     What  load  would  it  carry  if  12  ft.  long?     Use  Ritter's 
formula  and  assume  the  material  mild  steel. 

In  the  beams  in  the  previous  problems  no  consideration  has  been  taken 
of  their  lateral  stiffness.  In  practice  it  is  customary  when  using  a  maximum 
working  fiber  stress  to  limit  the  ratio  of  unsupported  beam  length  to  flange 
width  to  from  12  to  20.  Where  the  ratio  exceeds  these  numbers  the  allow- 
able working  fiber  stress  should  be  reduced.  This  is  made  necessary  by  the 
fact  that  in  a  vertically  loaded  beam  the  upper  flange,  being  subjected  to 
compression,  is  liable  to  fail  as  a  column  by  buckling  and,  therefore,  to 
secure  the  same  factor  of  safety,  the  working  stress  in  this  flange  must  be 
reduced.  There  are  several  ways  by  which  allowance  is  made  for  this  condi- 
tion, all  of  which  are  empirical  and  subject  to  criticism.  Attention  is  called 
to  the  reduction  of  the  working  fiber  stress  by  the  use  of  a  column  formula, 
as  in  the  Cambria  Handbook, 


3000  b 

f  =  allowable  stress  in  pounds  per  square  inch. 
/  =  length  between  lateral  supports  in  inches. 
b  =  width  of  flange  in  inches. 


388  GRAPHICS  AND   STRUCTURAL  DESIGN 

This  formula  is  for  a  maximum  desired  working  stress  of  16,000.  For 
limiting  values  other  than  16,000  Ibs.  reduce  the  maximum  fiber  stresses  by 
the  percentage  that  the  above  formula  reduces  16,000  Ibs. 

Another  method  is  based  upon  the  tests  of  long  beams  by  the  Pencoyd 
Iron  Works  and  adopted  by  the  Carnegie  and  Bethlehem  Steel  companies' 
handbooks.  It  assumes  the  maximum  allowable  fiber  stress  applicable  to 
spans  of  20  flange  widths  and  a  uniform  reduction  as  the  ratio  of  span  to 
flange  width  increases  until  a  beam  whose  span  equals  70  flange  widths 
is  reached  when  the  allowable  working  fiber  stress  is  limited  to  one-half 
that  used  in  the  first  case. 

Still  another  method  is  that  explained  in  Chapter  VII  of  this  book. 

61.  What  uniform  load  will  a  i5-in.  I  beam  weighing  42  Ibs.  per  ft.  carry 
on  a  23-ft.  span  if  unsupported  laterally,  and  if  the  maximum  working  fiber 
stress  for  a  beam  whose  span  is  20  flange  widths  is  limited  to  16,000  Ibs. 
per  sq.  in.?     Compare  the  results  by  the  methods  given  above. 

62.  What  central  load  can  be  carried  upon  a  Bethlehem  girder  beam  G-2O 
weighing  112  Ibs.  per  ft.  on  a  3O-ft.  span?     The  maximum  fiber  stress  for 
a  span  of  20  widths  is  16,000  Ibs.  per  sq.  in.     Compare  the  results  obtained 
by  the  methods  given. 

*  63.   A  lo-ton  crane  is  to  be  carried  across  a  span  of  30  ft.  on  rolled 
beams.     The  maximum  wheel  load  is  21,000  Ibs.,  the  wheel  base  10  ft.  6  ins. 
The  maximum  desired  working  stress  is  15,000  Ibs.  per  sq.  in.  reduced  to 
allow  for  ratio  of  span  to  flange  width.     Select  a  standard  Cambria  beam 
and  test  its  ability  to  resist  a  lateral  pull  of  one-tenth  of  the  live  load  divided 
between  the  two  wheels  acting  on  the  beam. 

*  64.   A  25-ton  crane  is  to  be  carried  across  a  36-ft.  span  on  rolled  beams. 
The  maximum  wheel  load  is  40,000  Ibs.;  the  wheel  base  is  10  ft.  6  ins.     All 
other  conditions  are  the  same  as  in  Problem  63,  excepting  that  a  Bethlehem 
Grey  Mill  section  with  a  wide  flange  is  to  be  used.     Determine  the  section. 

WEB  STRESSES 

In  addition  to  the  points  already  considered,  in  designing  beams  with 
heavy  concentrations  of  loads  or  very  short  spans  the  ability  of  the  web  to 
resist  vertical  shearing,  vertical  crippling  and  horizontal  shearing  must  also 
be  taken  into  account.  In  the  case  of  plate  girders  the  web  is  reinforced 
with  stiffening  angles  to  resist  this  crippling.  Rolled  steel  beams  are  rarely 
weak  in  vertical  or  horizontal  shear;  timber*  beams  should  be  carefully 
examined  for  horizontal  shear;  while  in  reinforced-concrete  construction 

*  Specifications  frequently  call  for  an  impact  allowance  of  25  per  cent  in  the 
case  of  traveling  cranes. 


PROBLEMS  389 

careful  designing  to  resist  the  web  stresses,  including  horizontal  shear  and 
diagonal  tension,  is  of  vital  importance. 

The  horizontal  shear  at  any  section  of  a  homogeneous  beam  is 

s  =  ~-  Zay. 
Iw 

s  =  horizontal  shear  on  section  per  square  inch. 
I  =  moment  of  inertia  of  entire  section. 
V  =  total  vertical  shear. 
w  =  width  of  web. 

Zay  =  statical  moment  of  the  area  of  the  cross  section  on  one  side  of  the 
neutral   axis,  or   summation  of   areas   multiplied  by  their  re- 
spective distances  from  the  neutral  axis. 
65.   Show  that  for  a  timber  beam  of  width  b  and  depth  d 


^  2bd 

66.  If  a  is-in.  I  beam  weighing  42  Ibs.  per  ft.  is  to  be  used  on  varying 
spans  and  a  fiber  stress  of  16,000  Ibs.  is  allowed  in  tension  and  12,000  Ibs. 
in  shear,  what  is  the  minimum  span  for  which  the  horizontal  shear  may  be 
neglected? 

67.  Allowing  a  tensile  fiber  stress  of  1500  Ibs.  per  sq.  in.  and  a  shearing 
fiber  stress  with  the  grain  of  120  Ibs.  per  sq.  in.,  what  uniform  load  can 
be  carried  by  a  yellow  pine  timber  12  ins.  deep,  3  ins.  wide  on  a  i2-ft.  span? 

68.  Allowing  a  tensile  fiber  stress  of  900  Ibs.  per  sq.  in.  and  a  shearing 
fiber  stress  with  the  grain  of  100  Ibs.  per  sq.  in.,  what  uniform  load  will 
a  spruce  timber  16  ins.  deep  X  4  ins.  wide  carry  on  a  i6-ft.  span? 

69.  What  is  the  maximum  span  upon  which  a  yellow  pine  timber  12  ins. 
deep  can  be.  used  to  carry  a  uniform  load  if  its  fiber  stress  is  to  be  limited 
to  1200  Ibs.  per  sq.  in.  and  its  deflection  to  yj^  of  its  span?     Assume  its 
modulus  of  elasticity  as  1,500,000  Ibs.  per  sq.  in. 

70.  What  central  load  will  a  white  oak  timber  16  ins.  deep  X  4  ins.  wide 
carry  on  an  i8-ft.  span  if  stressed  to  1500  Ibs.  per  sq.  in.?     What  would 
be  its  maximum  deflection   if  the  modulus  of  elasticity  were  taken  at 
1,500,000  Ibs.  per  sq.  in.? 

COMBINED  FLEXURE  AND  DIRECT  STRESS 

Sometimes  beams  are  subjected  to  direct  compression  or  tension  in  addi- 
tion to  flexural  stresses.  It  is  usually  sufficient  to  take  the  algebraic  sum 
of  the  flexural  and  direct  stresses  at  the  extreme  fibers  of  the  beam.  When 
greater  accuracy  than  this  is  required,  Johnson's  formula,  as  modified  by 


390  GRAPHICS  AND   STRUCTURAL  DESIGN 

Merriman,  can  be  used.  According  to  this  formula  the  fiber  stress  due  to 
bending,  when  direct  compression  or  tension  acts  upon  a  piece  in  addition 
to  bending,  is 

/'=        Me 
~  r.nPP 

*m  E 

The  —  is  for  a  compression  force. 
The  -}-  is  for  a  tensile  force. 

p 

The  combined  stress  is/'  ±  — 

A 

P  =  compression  or  tensile  force  in  pounds. 
/  =  span  in  inches. 

Simple  beam  uniform  load  =  —  =  —  . 

m      9.6 

Simple  beam  central  load=  —  =  -L. 

m      12 

A  =  area  of  cross  section  in  square  inches. 

An  exact  method  is  given  by  Merriman  but  its  application  is  too  difficult 
to  be  generally  used. 

71.  A  yellow  pine  beam  10  ft.  long  and  12  ins.  square  is  subjected  to  a 
compression  force  acting  along  its  length  of  50,000  Ibs.  while  carrying  a 
uniform  load  of  20,000  Ibs.  What  is  the  fiber  stress?  E  =  1,500,000. 

What  is  the  maximum  fiber  stress,  using  the  formula 

,=        Me 
~ 


72.  A  horizontal  steel  tension  bar  has  a  cross  section  of  9   X  i|  ins. 
It  is  subjected  to  a  unit  stress  in  tension  of  10,000  Ibs.  per  sq.  in.     The  bar 
is  20  ft.  long.     Determine  the  maximum  resulting  fiber  stress  due  to  the 
combined  action  of  its  weight  and  its  tensile  load.     The  bar  runs  vertically 
with  its  i^-in.  edge  down.    E  =  30,000,000. 

jtt  __  i_ 
m      9.6 

73.  In  Fig.  23  the  top  plate  is  20  X  I  in.,  the  sides  are  i5-lb.  channels 
at  40  Ibs.  per  foot  each.    Assume  the  gross  cross  section  loaded  with  7500 


PROBLEMS 


391 


Ibs.  per  sq.  in.     Chord  25  ft.  long.    E   =    30,000,000.    Determine  the 
maximum  fiber  stress  due  to  direct  loading  and  its  own  weight.     Use  formula 


E 


74.  In  Fig.  24  the  top  plate  is  24  X  |  in.  The  sides  are,  two  upper 
angles  4  X  4  X  •&  in.,  two  plates  24  X  }f  in.  and  two  bottom  angles 
6  X  4  X  1  in.  Area  of  combined  section  is  82.3  sq.  ins.  Inertia  of  sec- 
tion is  6969.  Assume  chord  20  ft.  long.  Uniform  load,  including  live  load 


FIG.  23. 


FIG.  24. 


FIG.  25. 


and  weight  of  section,  is  4000  Ibs.  per  foot  of  span.  Compression  load 
400,000  Ibs.  acting  along  the  center  of  gravity  of  the  section.  What  is  the 
total  maximum  fiber  stress?  E  =  30,000,000. 


i 
9.6 


75.  What  is  the  maximum  fiber  stress  in  Fig.  25?  Two  angles  4  X 
3  X  fV  in.,  one  plate  10  X  TVm-  Compression  due  to  direct  load  acting 
at  center  of  gravity  of  section  is  10,000  Ibs.  per  sq.  in.  on  gross  section. 
(In  this  problem  neglect  rivet  holes.)  Length  of  piece  8  ft.  o  ins.  Neg- 
lect weight  and  assume  a  central  load  due  to  purlin  of  1600  Ibs.  What  is 
the  maximum  fiber  stress? 

/'=        Me 


E  =  30,000,000. 


*L  —  JL 
m      12 


392  GRAPHICS  AND  STRUCTURAL  DESIGN 

RIVETING 

The  following  considerations  are  useful  in  designing  riveted  joints  for 
structural  work. 

1.  The  rivet  strength  is  calculated  upon  the  nominal  diameter  of  the 
rivet  notwithstanding  the  fact  that  the  hole  in  which  the  rivet  is  driven 
is  generally  Txg  in.  to  £  in.  greater  than  the  rivet. 

2.  The  unit  shearing  strength  of  steel  rivets  is  taken  about  £  the  unit 
tensile  strength  of  the  steel  and  the  unit  bearing  strength  of  rivets  is  taken 
at  double  their  unit  shearing  strength. 

3.  Rivets  should  not  be  used  in  tension. 

4.  Use  table  of  rivet  values. 

76.  A  24-in.  I  beam  weighing  80  Ibs.  per  ft.  carries  a  uniform  load  on  a 
i9-ft.  span.     First  find  what  load  it  will  carry  if  the  extreme  fiber  stress  is 
16,000  Ibs.  per  sq.  in.  and  then  find  how  many  f-in.  rivets  will  be  required 
to  secure  two  4  X  4  X  f-in.  angles  to  the  web  of  the  beam  and  transfer 
the  reaction  due  to  the  given  load.     Compare  your  result  with  the  stand- 
ard framing.     Allow  10,000  Ibs.  per  sq.  in.  shear. 

77.  Two  4  X  4  X  rVm-  angles  are  fastened  back  to  back  to  a  f-in.  plate. 
How  many  f-in.  rivets  will  be  required  in  the  angles  if  the  net  section 
of  the  material  is  stressed  12,000  Ibs.  per  sq.  in.?    Allow  one  rivet  hole 
if  in.  diam.  through  angles  and  plate.     Allow  7500  Ibs.  per  sq.  in.  in  shear. 

78.  The  long  leg  of  a  5X  35  X  f-in.  angle  is  fastened  to  a  f-in.  plate. 
Assume  one  rivet  hole  it  in.  in  diameter,  in  a  section.     Allow  a  fiber  stress 
of  12,000  Ibs.  per  sq.  in.  on  the  net  section  and  determine  how  many  f-in. 
rivets  are  required.    Allow  7500  Ibs.  per  sq.  in.  in  shear. 

TIMBER  COLUMNS 

The  U.  S.  Dept.  of  Agriculture  has  made  a  large  number  of  tests  upon 
timbers  and  the  following  is  their  formula  suggested  for  timber  columns. 


700+  isc  +  c2 

To  find  the  safe  load  divide  p  by  the  required  factor,  where 

p  =  ultimate  strength  in  pounds  per  square  inch. 
F  =  crushing  strength  of  timber. 

_  _L  Length  (inches) 

d     small  diameter  (inches) 

79.  What  load  can  be  safely  carried  by  a  yellow  pine  post  10  ins.  square 
and  20  ft.  high?  Use  a  factor  of  4  and  consider  the  ultimate  crushing 
strength  5000  Ibs.  per  sq.  in. 


PROBLEMS  393 

80.  A  cedar  post  is  22  ft.  high  and  is  10  X  12  ins.  in  cross  section.    What 
load  can  it  safely  carry  if  its  ultimate  crushing  strength  is  assumed  at  3500 
Ibs.  and  a  factor  of  safety  of  5  is  desired? 

81.  A  spruce  column  is  18  ft.  high  and  9  ins.  square.     What  load  can  it 
safely  carry  if  its  ultimate  crushing  strength  per  sq.  in.  is  taken  as  4000  Ibs. 
and  a  factor  of  safety  of  4  is  desired? 

PLATE  GIRDERS 

In  designing  girders  of  greater  depth  than  rolled  sections  (24  and  36  ins.) 
the  following  method  is  easier  and  is  sufficiently  accurate.     In  Fig.  26, 
A  =  area  of  one  flange  in  square  inches. 

h  =  distance  between  the  centers  of  gravity  of  the  two  flanges  in 
inches.  Where  the  flange  is  made  up  of  angles  and  plates 
this  distance  is  frequently  assumed  as  the  distance  back  to 
back  of  the  angles. 

/  =  mean  fiber  stress  in  the  flange  in  pounds  per  square  inch. 
M  =  external  bending  on  the  section  measured  in  inch  pounds. 

Then  M  =  A  X/Xh. 

This  formula  assumes  the  web  as  resisting  shear  only.     When  the  web  is 
assumed  to  resist  bending  also  the  formula  becomes 


Here  a  is  the  area  of  the  web  plate  in  square  inches. 

82.  Derive  the  two  formulae  for  girders, 

M  =  AxfXh    and    M 

83.  Find  the  net  flange  area  required  at  the  middle  of  the  following  girder. 
Span  60  ft.    Depth  of  girder  back  to  back  of  flange  angles  is  6  ft.  o  ins. 
Uniform  dead  load  is  600  Ibs.  per  ft.  (this  covers  ties,  rails  and  metal  of  girder, 
etc.).     Uniform  live  load  is  2250  Ibs.  per  lineal  foot.    Increase  the  live-load 
bending  80  per  cent  to  allow  for  impact  due  to  a  moving  train  load.    Allow  a 
fiber  stress  of  16,000  Ibs.  per  sq.  in.    Assume  that  the  web  takes  shear  only. 

84.  A  girder  spans  80  ft.     Its  depth,  back  to  back  of  flange  angles,  is  7  ft. 
o  ins.    Uniform  dead  load  is  700  Ibs.  per  ft.  (this  covers  ties,  rails,  metal  of 
girder,  etc.).     Uniform  live  load  is  2000  Ibs.  per  lineal  foot.    Increase  the 
live-load  bending  80  per  cent  to  allow  for  impact  due  to  a  moving  train 
load.    Allow  a  fiber  stress  of  16,000  Ibs.  per  sq.  in.       Assume  that  one- 
eighth  of  the  web  acts  as  flange  area  and  that  the  web  is  f  in.  thick.    What 


394 


GRAPHICS  AND   STRUCTURAL  DESIGN 


additional  area  is  required  to  complete  the  net  flange  section  30  ft.  from  the 
left  abutment? 

85.  A  girder  which  spans  50  ft.  is  to  be  made  5  ft.  o  ins.  deep,  back  to 
back  of  flange  angles.  It  carries  two  wheel  loads  12  ft.  o  ins.  center  to 
center  of  80,000  Ibs.  each.  Assume  a  dead  load  including  girder  weight  of 
200  Ibs.  per  ft.  Increase  the  live-load  bending  20  per  cent  to  cover  impact. 
Allow  a  fiber  stress  of  16,000  Ibs.  per  sq.  in.  Assume  the  web  T\  in.  thick 
and  that  one-eighth  of  it  is  considered  as  being  flange.  Determine  what 
additional  area  is  required  to  complete  the  net  flange  area  at  the  point  of 
maximum  bending. 

f^ — 22-- 

' 1 


FIG.  26. 


FIG.  27. 


FIG.  28. 


86.  A  bridge  which  spans  60  ft.  is  to  be  built  for  a  crane.  Assume  that 
each  girder  weighs  15,000  Ibs.  and  that  this  forms  a  uniform  load.  The 
trolley  wheels  are  6  ft.  apart  and  each  wheel  load  is  16,000  Ibs.  The  girder 
at  the  center  is  48  ins.  deep,  back  to  back  of  angles.  On  account  of  stiffen- 
ing the  crane  laterally  the  flange  plates  are  made  22  ins.  wide.  Assume  the 
section  like  Fig.  27.  Assume  the  web  as  taking  shear  only.  The  fiber  stress 
in  the  compression  flange  is  to  be  9000  Ibs.,  while  that  in  the  tension  flange 
can  be  12,000  Ibs.  per  sq.  in.  Determine  the  dimensions  of  the  plates. 

SHAFTING 

Considering  the  differential  area,  Fig.  28,  the  fiber  stress  is  proportional 
to  its  distance  r  from  the  center  so  that  if  p  is  the  fiber  stress  at  a  distance 
R  from  the  center  the  stress  pr  at  this  distance  is 


Pr  = 


The  force  acting  on  an  area  dA  then  is  dF  =  pr  X  dA, 
The  moment  of  this  force  about  the  center  then  is 


dMt 


Integrating 


PROBLEMS  395 

It  will  be  noticed  that  this  is  in  the  form  of  the  general  equation  for 
torsion. 


where  /  is  the  polar  moment  of  inertia. 

87.  Allowing  a  limiting  shearing  working  stress  of  7000  Ibs.  per  sq.  in. 
upon  a  steel  round  2  ins.  in  diameter,  what  twisting  moment  measured  in 
inch  pounds  will  the  round  carry?    What  would  be  the  force  applied  at  a 
radius  of  i  ft.? 

88.  Allowing  a  limiting  shearing  working  stress  of  7000  Ibs.  per  sq.  in., 
what  twisting  moment  in  inch  pounds  would  be  carried  by  a  2  in.  square 
shaft?    Acting  at  a  radius  of  i  ft.  what  force  would  this  correspond  to? 

89.  Find  the  ratio  of  the  twisting  moments  carried  by  a  round  shaft 
and  a  square  shaft  of  the  same  sectional  area  and  the  same  limiting  fiber 
stress. 

90.  What  horse  power  will  a  shaft  of  diameter  d  transmit  when  making 
.V  revolutions  per  minute  and  subjected  to  a  working  fiber  stress  of  p  Ibs. 
per  sq.  in.  at  its  circumference? 

TORSIONAL  DEFLECTION 

It  is  sometimes  desirable  to  limit  the  torsional  deflection  of  a  shaft,  in 
which  case  the  angle  of  deflection  in  degrees  is  given  by 

A=    *«XL. 

1660  X  d* 

Mt  =  twisting  moment  in  inch  pounds. 
L    =  length  of  shaft  in  feet. 
d     =  diameter  of  shaft  in  inches. 

91.  A  shaft  2  ins.  in  diameter  and  60  ft.  long  is  subjected  to  a  twisting 
moment  which  produces  an  extreme  fiber  shearing  stress  of  6000  Ibs.  per 
sq.  in.    Through  what  angle  is  one  end  of  the  shaft  twisted  ahead  of  the 
other? 

COMBINED  BENDING  AND  TWISTING 

Shafting  must  generally  be  designed  for  combined  torsion  and  bending. 
There  are  several  formulae  but  according  to  Guest's  law  the  following  may 
be  used: 


Here  M  e  =  equivalent  bending  moment. 

M  =  bending  moment. 
T  =  twisting  moment. 


GRAPHICS  AND   STRUCTURAL  DESIGN 


Having  found  the  equivalent  bending  moment,  the  shaft  diameter  can  be 
determined  by  placing  these  values  in  the  equation 

M  =  ^- 

e 

I  =  moment  of  inertia  referred  to  axis. 

e  =  distance  from  center  of  gravity  to  extreme  fibers. 

92.  In  Fig.  29  a  force  of  2000  Ibs.  acts  on  the  teeth  of  a  pinion  5  ins.  in 
diameter.  The  distance  from  the  center  line  of  the  pinion  to  the  center  of 
the  adjacent  bearing  is  6  ins.  What  diameter  of  shaft  will  be  required  if 
oooo  Ibs.  per  sq.  in.  is  permitted  in  flexure? 


~^3\ 

1 

FIG.  29. 


Jpp^VXu^^W^ 

TJ^V  >p:L:^  ^ 


FIG.  31. 


FIG.  32. 


93.  In  Fig.  30  a  force  of  2500  Ibs.  acts  upon  the  teeth  of  a  pinion  4%  ins. 
in  diameter.  The  pinion  is  inside  the  bearings  as  shown.  Using  a  flexural 
fiber  stress  of  9000  Ibs.  per  sq.  in.  what  diameter  of  shaft  is  required? 

STRESSES  IN  STRUCTURES  DETERMINED  ALGEBRAICALLY 

In  Fig.  31  to  determine  the  stress  in  any  piece,  as  FA,  cut  the  truss  at  a 
section  containing  this  piece  and  then  equate  the  internal  and  external 
moments  about  any  point  that  most  conveniently  gives  the  desired  result. 
Consult  Chapter  IV. 

In  Fig.  32  BC  =  2400  Ibs.,  CD  =  DE  =  EF  =  etc.  =  4800  Ibs. 

94.  Calculate  algebraically  the  stress  in  GC  and  AG  in  Fig.  32. 

95.  Determine  algebraically  the  stresses  in  GH  and  HD  in  Fig.  32. 

96.  Calculate  the  stresses  in  HI  and  I A  in  Fig.  32. 

97.  Calculate  the  stresses  in  PL  and  LM  in  Fig.  32. 
Problems  94  to  97  should  be  checked  graphically. 

In  Fig.  33  BC  =  CD  =  DE  =  EE',  etc.  =  10,000  Ibs. 

98.  Calculate  algebraically  the  stresses  in  BF  and  FA  in  Fig.  33. 

99.  Calculate  algebraically  the  stresses  in  FG  and  GC  in  Fig.  33. 
100.   Calculate  algebraically  the  stresses  in  GH  and  HI  in  Fig.  33. 
Problems  98  to  100  should  be  checked  by  method  of  coefficients. 


PROBLEMS 
GRAPHICAL  ANALYSIS  or  TRUSS  STRESSES 


397 


1 01.  In  using  the  common  graphical  method  of  determining  truss  stresses, 
what  assumption  is  made  concerning  the  construction  of  the  truss?  What 
are  the  three  conditions  necessary  for  equilibrium  in  a  structure  acted  on  by 
forces? 


M/    t \|  t 


FIG.  33. 


FIG.  34. 


102.  What  is  a  force  triangle?    A  force  polygon?    What  can  you  say 
concerning  the  force  polygon  representing  a  number  of  concurrent  forces 
in  equilibrium?    What  general  direction  will  the  forces  in  the  polygon  take? 

103.  Make  a  sketch  of  any  simple  truss  and  show  how  force  polygons 
may  be  used  to  determine  the  magnitude  and  character  of  the  stresses  in  the 
structures  when  the  external  forces  are  known. 

104.  Take  Fig.  34,  letter  the  truss  and  make  a  careful  free-hand  sketch 
of  the  stress  diagram,  then  find  the  character  of  the  stresses. 


FIG.  35. 


FIG.  36. 


FIG.  37. 


105.  Take  Fig.  35,  letter  the  truss  and  make  a  careful  free-hand  sketch 
of  the  stress  diagrams,  first  with  a  live  load  L  at  i  and  then  at  2.     Deter- 
mine the  character  of  the  stress.  * 

1 06.  What  is  an  equilibrium  polygon?    When  a  number  of  noncon- 
current  forces  are  in  equilibrium  what  graphical  conditions  are  fulfilled? 

107.  In  Fig.  36  show  by  a  free-hand  sketch  how  the  resultant  of  the 
three  forces  acting  on  the  piece  can  be  determined  in  direction,  magnitude 
and  located  in  relation  to  the  piece. 

1 08.  In  Fig.  37  show  by  a  free-hand  sketch  how  the  reactions  due  to  the 
three  parallel  forces  acting  on  the  beam  can  be  determined. 

109.  In  Fig.  38  a  truss  is  acted  on  by  wind  forces.    Both  ends  of  the 
truss  are  fixed.     Show  how  the  reactions  can  be  found. 

no.  In  Fig.  39  the  truss  is  acted  on  by  wind  pressure.    The  right-hand 
end  is  on  rollers  while  the  other  end  is  fixed.    Show  by  a  free-hand  sketch 


398 


GRAPHICS  AND  STRUCTURAL  DESIGN 


how  the  magnitude  of  the  right  reaction  and  the  magnitude  and  direction 
of  the  left  reaction  can  be  found. 

Explain  how  an  equilibrium  polygon  may  be  used  as  a  bending- 


in. 


moment  diagram  and  prove  that  the  statement  is  correct. 

112.  Make  a  free-hand  sketch  showing  by  an  equilibrium  polygon  how 
the  bending  can  be  found  on  the  beam  in  Fig.  40. 


FIG.  38. 


FIG.  39. 


FIG.  40. 


The  equilibrium  polygon  may  also  be  used  in  determining  the  stresses  as 
follows:  In  finding  truss  stresses  algebraically  the  internal  and  external 
moments  were  equated.  The  number  of  calculations  can  be  reduced  by 
using  an  equilibrium  polygon  for  the  external  bending  moment,  the  single 
diagram  serving  for  all  points. . 

113.  In  Fig.  41  show  how  the  stresses  AI,  HI  and  HE  may  be  deter- 
mined by  combining  both  graphical  and  algebraic  methods  as  suggested. 

114.  In  Fig.  42  determine  the  dead-load  stresses  in  members  AJ,  JK 
and  IF.    The  bridge  has  the  following  dimensions:  Span,  150  ft.;  height, 
center  to  center  chords,  30  ft.;  uniform  dead  load,  2000  Ibs.  per  ft.,  carried 
at  lower  apex  points. 


FIG.  41. 


FIG.  42. 


FIG.  43- 


115.  Show  how  a  bending-moment  diagram  may  be  determined  graphi- 
cally to  give  the  maximum  moments  on  all  sections  of  a  girder  for  two  loads, 
a  constant  distance  a  apart,  moving  across  the  girder. 

1 1 6.  Show  how  a  bending-moment  diagram  may  be  determined  graphi- 
cally to  give  the  maximum  moments  on  all  sections  of  a  girder,  due  to  a 
number  of  loads  (use  6),  a  constant  distance  apart,  moving  across  the  girder. 

117.  Show  how  to  construct  a  diagram  giving  the  maximum  shears  at 
all  points  on  a  girder  due  to  a  locomotive  and  train  load  passing  across  it. 
Prove  that  the  diagram  gives  the  required  results. 


PROBLEMS 


399 


1 1 8.  In  Fig.  43  show  how  the  maximum  stresses  due  to  moving  wheel 
loads  crossing  the  bridge  may  be  determined  for  members  AH,  HI  and  //. 
Use  a  combined  graphical  and  algebraic  method. 

STRESSES  IN  CRANE  FRAMES 

Show  how  to  determine  the  stresses  in  the  following  crane  frames.  Ex- 
plain the  methods  fully. 

1.  Find  direct  stresses  due  to  live  load  at  maximum  radius. 

2.  Find  direct  stresses  due  to  live  load  at  minimum  radius. 

3.  Draw  bending-moment  diagrams  for  members  subjected  to  bending. 
If  necessary  draw  a  stress  diagram  for  position  of  live  load  producing 
maximum  bending  iri  any  member. 

4.  Draw  stress  diagram  for  dead  load. 


FIG.  44. 


FIG.  45. 


FIG.  46. 


1 19.  Make  the  necessary  stress  and  moment  diagrams  for  the  frame  given 
hi  Fig.  44. 

1 20.  Make  the  necessary  stress  and  moment  diagrams  for  the  frame  given 
in  Fig.  45. 

121.  Make  the  necessary  stress  and  moment  diagrams  for  the  frame  given 
in  Fig.  46. 


FIG.  47. 


FIG.  48. 


122.  Make  the  necessary  stress  and  moment  diagrams  for  the  frame 
given  in  Fig.  47,  taking  the  chain  or  rope  pull  into  account. 


400  GRAPHICS  AND  STRUCTURAL  DESIGN 

123.  Make  the  necessary  stress  and  moment  diagrams  for  the  frame 
given  in  Fig.  48,  taking  the  chain  or  rope  pull  into  account. 

124.  In  Fig.  48  show  how  to  stiffen  FB  so  that  the  carriage  can  travel 
back  close  to  the  post.    Also  show  how  channels  used  for  EA  could  be 

stiffened  to  reduce  the  -  value  and  permit  the  load  chain  to  travel  back 
close  to  the  post. 

PLATE-GIRDER  DESIGN 

Consult  Chapter  XI  for  information  concerning  these  problems. 

125.  Assume  a  half  dozen  wheel  loads  and  their  distances  apart  and 
explain  how  the  diagram  of  maximum  bending  moments  would  be  drawn. 
Show  how  to   include   the   bending   due   to  the  girder  weight  in  this 
diagram. 

126.  A  girder  has  a  span  of  65  ft.    Assume  that  it  carries  a  load  equiva- 
lent to  a  uniform  load  of  2000  Ibs.  per  ft.     Calculate  the  bending  at  the 
center  and  draw  the  bending-moment  diagram. 

LENGTHS  or  FLANGE  PLATES 

In  plate  girders  the  flange  force  varies  from  zero  at  the  supports  to  a 
maximum  near  the  center  of  a  girder.  If  the  girder  is  of  approximately 
constant  depth  the  bending-moment  diagram  can  represent  the  flange  force 
by  changing  the  scale  ;  thus, 

M  =  A  XpXh 
and  the  flange  force 


h  being  constant  F  varies  as  M.    This  diagram  can  now  be  used  to  deter- 
mine the  lengths  of  the  several  plates  and  angles  composing  the  flange. 

127.  Assume  any  girder.    Show  how  to  determine  the  lengths  of  the 
several  flange  plates. 

FLANGE  RIVETING 

128.  The  diagram  representing  the  forces  acting  in  the  girder  flanges 
can  be  used  to  determine  the  riveting  of  the  flange  angles  to  the  web  and 
also  the  several  flange  plates  together.     Show  how  to  do  this. 

129.  A  girder  spans  60  ft.    Assume  the  depth  back  to  back  of  angles  as 
6  ft.    The  flange  at  the  middle  is  composed  of  two  6  X  6  X  f-in.  angles, 
and  two  1  5  X  f-in.  plates.    The  web  is  f  in.  thick,  rivets  are  £  in.  in  diameter. 


PROBLEMS  401 

Assume  the  bending,  including  dead  load,  live  load  and  impact,  as  covered 
by  a  uniform  load  on  the  girder  of  6000  Ibs.  per  ft.  of  span.  By  the  method 
suggested  in  Problem  128,  find  the  number  of  rivets  required  between  the 
web  and  flange  angles  for  the  panel  extending  from  6  ft.  to  12  ft.  from  the 
left  pier.  Allow  11,000  Ibs.  shear  and  22,000  Ibs.  bearing  on  the  rivets. 

130.  Take  the  same  data  as  in  Problem  129,  and  determine  the  number 
of  rivets  in  the  panel  extending  from  24  ft.  to  30  ft.  from  the  left  pier. 
Assume  that  the  web  resists  bending. 

131.  Explain  how  to  construct  a  diagram  of  maximum  shears  for  a 
locomotive  and  train  load  passing  across  a  bridge. 

132.  Estimate  the  reaction  from  the  data  given  in  Problem  129  (this 
includes  dead  load,  live  load  and  impact)  and  determine  the  shearing  fiber 
stress  in  the  web  plate  which  is  6  ft.  deep  and  f  in.  thick. 

133.  Explain  the  object  of  web  stiffeners.    What 
is  the  usual  spacing  of  web  stiffeners  in  ordinary 
railroad  practice?    How  thick  must  the  web  be,  in 
relation  to  the  girder  depth,  to  permit  the  omission 
of  stiffeners? 

134.  Show  by  means  of  a  diagram  of  maximum 
shears  that  the  maximum  vertical  shear  at  any 


o    o    o 


s  U- 
t 


point   due    to   a   uniform    load   occurs   when   the  FIG.  49 

girder    is    loaded    from    this    point    to    the    more 
remote  of  the  two  reactions. 

Instead  of  the  method  previously  indicated  for  obtaining  the  rivet  spacing 
between  the  flange  and  the  web,  the  following  is  the  method  more  generally 
used.  In  Fig.  49 

Rxh 
V 

s  =  rivet  pitch,  inches. 
R  =  rivet  value,  pounds. 

h  =  distance  between  gauge  lines  of  flange  angles,  inches. 
V  =  vertical  shear,  generally  taken  as  the  mean  of  the  panel  in 
which  the  rivets  are  to  be  spaced. 

135.  Assume  the  same  general  data  as  given  in  Problem  129.     Use  the 
method  just  given  and  determine  the  rivet  spacing  in  the  lower  flange  for  the 
section  of  the  girder  extending  from  12  ft.  to  18  ft.  to  the  right  of  the  left 
support.    Assume  that  the  web  takes  shear  only.    Assume  that  h  =  64!. 
Use  f-in.  rivets,  allowing  11,000  Ibs.  in  shear  and  22,000  Ibs.  in  bearing. 

136.  Given  f-in.  web,  6  X  4  X  jVm-  flange  angles,  h  =  55  ins.    Rivet 
diameter  £  in.,  shearing  stress  10,000  Ibs.  per  sq.  in.,  bearing  stress  20,000 


402  GRAPHICS  AND  STRUCTURAL  DESIGN 

Ibs.  per  sq.  in.,  vertical  shear  75,000  Ibs.  What  rivet  spacing  is  required 
at  this  section?  Assume  that  the  web  takes  shear  only. 

When  concentrated  loads  are  transferred  to  the  web  plates  through  the 
flange  angles  the  resultant  of  the  vertical  and  horizontal  forces  must  be 
found  and  the  rivet  spacing  determined  from  this  resultant.  These  con- 
centrations are  usually  assumed  as  distributed  over  a  certain  distance  along 
the  flange,  say  36  ins. 

The  change  in  the  horizontal  flange  force  per  inch  of  flange  length  is 


The  vertical  force  per  inch  of  flange  length  is 
e  load 


distance  distributed 

T> 

The  resultant  fr  =  Vtf  +  /22.      Rivet  spacing  =  — . 

Jr 

137.  Solve  Problem  129,  taking  into  account  the  load  of  6000  Ibs.  per  foot 
carried  from  the  flange  angles  to  the  web  by  the  rivets. 

138.  Given  web  plates  f  in.    Angles  6  X  6  X  f  in.    Rivets  £  in.  in  diameter. 
Shearing  stress  11,000  Ibs.    Bearing  stress  22,000  Ibs.    Total  vertical  shear 
at  middle  of  panel  under  consideration  200,000  Ibs.    h  =  69  ins.    Assume 
load  of  45,000  Ibs.  distributed  over  36  ins.  of  upper  flange.    (This  includes 
dead  load,  live  load  and  impact.)     Determine  the  rivet  spacing.    Assume 
that  the  web  takes  shear  only. 

The  horizontal  shear  between  the  flange  and  the  web  is  reduced  when  the 
web  is  assumed  as  assisting  in  carrying  the  bending.  The  change  in  this 
shear  transferred  through  the  rivets  depends  upon  the  ratio 


where 

A  —  net  area  of  flange  angles  and  plates, 

a  =  gross  web  area 
A 


n 

and,  as  before,  rivet  spacing  =  — 

Jr 


PROBLEMS 


403 


139.  In  Problem  138  assume  that  one-eighth  of  the  web  acts  with  the 
flange  and  that  at  this  point  the  net  flange  area  is  12  sq.  ins.,  not  including 
one-eighth  of  the  web.     Find  the  required  rivet  spacing. 

140.  Determine  the  rivet  spacing  satisfying  the  following  conditions: 
Vertical  shear  160,000  Ibs.;  h  =  64.75  ins-    Net  flange  area,  not  including 
one-eighth  of  web,  20  sq.  ins.;  web  72  X  f  in.     Upper  flange  concentration 
45,000  Ibs.  acting  over  36  ins.     Shearing  stress  11,000  Ibs.  and  bearing 
stress  22,000  Ibs.  per  sq.  in.     Use  rivets  f-in.  in  diameter. 


*  5  x.s 


ol'° 
ojjo 
oo 

0,o 


FIG.  50. 


FIG.  51. 


FIG.  52. 


141.  A  girder  running  over  two  supports  is  cantilevered  5  ft.  beyond  one 
of  them  and  carries  a  concentrated  load  at  its  end,  producing  an  average 
fiber  stress  in  the  flange  angles  of  15,000  Ibs.  per  sq.  in.     Assume  the  section 
given  in  Fig.  50  and  determine  the  spacing  in  the  flange  angles  of  the  canti- 
levered portion  for  f-in.  rivets.    Allow  10,000  Ibs.  per  sq.  hi.  in  shear  and 
double  this  in  bearing. 

142.  Fig.  51  is  made  up  of  1-8  in.  LJ  ii£  Ibs.  per  ft.,  area  3.35  sq.  in. 
and  i-i2-in.  /3i£  Ibs.  per  ft.,  area  9.26  sq.  ins.    The  average  upper  flange 
stress  is  8000  Ibs.  at  the  middle  of  a  22-ft.  span,  carrying  a  central  load. 
What  rivet  spacing  should  be  used  in  securing  the  channel  to  the  beam  if 
the  rivets  are  £  in.,  the  shearing  fiber  stress  10,000  Ibs.  per  sq.  in.  and 
the  bearing  stress  double  this? 

143.  In  Problem  142  determine  the  rivet  spacing  4  ft.  from  the  left 
support  if  the  beam  carries  a  uniform  load  instead  of  a  central  one?    What 
is  the  rivet  spacing  at  this  point? 

WEB  SPLICE 

When  the  web  is  assumed  as  resisting  bending  it  becomes  necessary  to 
rivet  the  splice  plates  so  that  they  and  the  riveting  shall  replace  the  broken 
section  in  bending  as  well  as  in  shear.  See  Chapter  XI. 

144.  A  girder  is  5  ft.  deep  and  has  a  f^-in.  web  plate;  one-eighth  of  the 
web  is  assumed  as  acting  as  flange;  the  average  flange  stress  acts  29  ins.  from 
the  neutral  axis  and  is  12,000  Ibs.  per  sq.  in.    Using  two  splice  plates,  how 


404 


GRAPHICS  AND  STRUCTURAL  DESIGN 


thick  must  they  be  made  if  they  can  be  only  50  ins.  deep?  The  fiber  stress 
in  the  splice  plates  is  to  equal  that  in  the  web  plate  at  the  same  distance 
from  the  neutral  axis  of  the  girder. 

145.  Assume  the  data  the  same  as  that  given  in  Problem  144  and  Fig.  52. 
How  many  vertical  rows  of  rivets  will  be  required  to  make  the  web  splice 
properly  carry  its  portion  of  the  moment?    Allow  10,000  Ibs.  per  sq.  in.  in 
shear  and  20,000  Ibs.  in  bearing. 

146.  A  girder  6  ft.  deep  has  a  |-in.  web  plate  which  is  to  be  spliced  at  the 
middle.     How  thick  must  each  of  two  splice  plates  be  if  they  can  be  made 
only  58 1  ins.  deep?  The  fiber  stress  at  the  backs  of  the  angles  is  12,000  Ibs. 
and  the  fiber  stress  in  the  splice  plates  is  to  equal  the  fiber  stress  in  the  cor- 
responding position  in  the  web  plates. 

147.  Take  the  girder  in  Problem  146  and  Fig.  53.    How  many  rows  of 
rivets  will  be  required,  allowing  12,000  Ibs.  per  sq.  in.  shear  and  24,000  Ibs. 
in   bearing?    Design  the  splice   to   resist  bending.     Use  £-in.  diameter 
rivets. 


R=.15000*        R 


FIG.  53. 


FIG.  54. 


148.  Assume  that  a  girder  6  ft.  deep  and  6o-ft.  span  carries  a  moving 
load  of  6000  Ibs.  per  ft.     Determine  the  maximum  shear  at  the  middle  and 
design  the  splice  plates,  assuming  that  the  web  takes  shear  only.    How 
many  f-in.  rivets  carrying  12,000  Ibs.  per  sq.  in.  shear  and  24,000  Ibs.  per 
sq.  in.  bearing  will  be  required,  the  web  being  f  in.  thick?     Depth  of  splice 
plates  is  58 J  ins. 

149.  The  lateral  bracing,  Fig.  54,  in  a  deck  girder  is  placed  in  the  plane 
of  the  upper  flanges.    Assume  that  the  tension  pieces  take  the  load.    The 
wind  load  is  assumed  as  300  Ibs.  per  ft.,  to  account  for  wind  blowing  on  the 
train,  and  30  Ibs.  per  sq.  ft.  of  girder.    The  combined  load  is  assumed  as 
acting  upon  the  upper  flange  through  the  rail,     (i)  Determine  the  apex 
loads  and  (2)  make  a  stress  diagram  showing  how  the  forces  acting  in  the 
wind  braces  may  be  determined.    Assume  vertical  girders  6  ft.  deep. 

150.  Take  the  data  given  in  Problem  149  and  determine  the  stresses  in 
the  bracing  algebraically. 


PROBLEMS 


405 


151.  Assume  that  the  lateral  truss  in  line  with  the  upper  flanges  trans- 
mits all  the  wind  forces  in  this  plane  to  the  ends  of  the  girder,  and  that  from 
here  the  forces  are  transferred  to  the  piers  by  the  end  bracing,  Fig.  55. 
Find  the  reactions  from  Problem  149.  Assume  that  %  R  goes  to  2,  the  other 
\  R  through  i  and  then  to  3.  Determine  the  forces  in  i,  2  and  3,  and 
design  them,  assuming  struts  with  fixed  ends  and  that  /  =  |  the  diagonal. 
Allow  a  fiber  stress  of  16,000  Ibs.  per  sq.  in. 


FIXED  AND  CONTINUOUS  BEAMS 

Fixed  and  continuous  beams  do  not  have  much  application  in  structural 
work;  however,  owing  to  the  monolithic  character  of  concrete  work  the 
bending  moments  are  frequently  taken  at  values  between  those  for  supported 
and  fixed  beams. 

152.  Fig.  56  is  a  beam  fixed  at  one  end  and  supported  at  the  other;  load 
is  at  the  middle  of  the  span.  The  point  of  inflection  is  at  X  =  f  /.  Show 
how  to  find  the  bending-moment  diagram  graphically. 


w 


FIG.  56. 


FIG.  57. 


FIG.  58. 


153.  A  beam  fixed  at  one  end  and  supported  at  the  other  carries  a  uniform 
load.     The  reaction  at  the  fixed  end  is  f  the  load.     Determine  the  bending 
moments  graphically  and  locate  the  points  of  inflection. 

154.  The  beam  in  Fig.  57  carries  a  uniform  load  W  Ibs.  per  ft.  over 
its  length  (/  +  2x).     Determine  graphically  the  value  of  x  in  terms  of  / 
so  that  the  central  span  shall  fulfill  the  conditions  of  a  fixed  beam,  i.e., 


WP- 
- 

24 


at  the  middle,  and  at  the  supports 


if-  -HZ. 

12 

155.  A  fixed  beam,  Fig.  58,  carries  a  central  load  W.  The  bending  at 
the  supports  is  —  M,  while  for  the  middle  the  bending  is  +  M.  (Numeri- 
cally equal  but  of  opposite  signs.)  Determine  graphically  these  moments 
and  the  points  of  inflection. 


406  GRAPHICS  AND   STRUCTURAL  DESIGN 

156.  If  a  beam  carries  a  uniform  load  and  is  restrained  at  the  supports 
so  that  the  bending  at  the  middle  is  T\j-  wP,  determine  graphically  the  bend- 
ing moment  at  the  supports  and  the  points  of  inflection. 

REINFORCED-CONCRETE  DESIGNING 

For  the  explanation  of  the  theory  and  derivation  of  the  formulae  see 
Chapter  XIV.    Also  consult  page  193  for  the  nomenclature. 

157.  Derive  the  following  formulae: 


A 
and 


-v/; 


158.  Design  a  rectangular  beam  to  carry  a  load  of  800  Ibs.  per  ft.  of  a 

WL 

2o-ft.  span.    Assume  M  =  --    Allow  /,  =  16,000  Ibs.,  fc  =  500   Ibs. 

10 

Tf> 

Take  d  =  20  ins.  and  assume  —*  =  15.    Determine  the  width  and  the  area 

£c 

of  the  reinforcing  steel. 

159.  What  size  rectangular  beam  and  reinforcement  will  be  required 
for  a  span  of  16  ft.    The  beam  ends  are  merely  supported  and  the  load 

is  1000  Ibs.  per  ft.  of  span.     Given  M  =  ——,    —  *  =  15,   tensile  stress  in 

o        H,c 

steel  16,000  Ibs.  per.  sq.  in.,  compression  in  concrete  500  Ibs.  per  sq.  in. 
and  the  width  of  the  beam  is  to  be  40  per  cent  of  its  effective  depth. 

1  60.   Derive  the  following  approximate  -formulae  for  rectangular  beams, 
the  symbols  having  the  significance  given  on  page  193. 

M,  =  £  X  A  Xfs,        Mc  =  l/c  X  b  X  (P. 

161.  The  purlins  on  a  roof  are  6  ft.  center  to  center.     How  thick  must  a 
reinforced  roof  slab  be  made  if  the  total  live  and  dead  load  is  80  Ibs.  per 

sq.  ft.?    Assume-—*  =  15;  fe=  500;  /•  =  14,000  and  M  =  —  • 
zic  o 

162.  A  reinforced-concrete  floor  carrying  a  combined  live  and  dead  load 
of  400  Ibs.  per  sq.  ft.  is  placed  over  steel  beams  that  are  on  6-ft.  centers. 

Taking  |=I5)    M  =  ^, 

allowable  compression  in  concrete  500  Ibs.  per  sq.  in.,  and  tension  in  steel 
14,000  Ibs.  per  sq.  in.,  find  depth  of  slab  and  area  of  reinforcement. 


PROBLEMS 


407 


163.  A  rectangular  tank  supported  on  its  lower  edges  is  5  ft.  wide,  6  ft. 
high  and  13  ft.  long  (inside  dimensions).    How  thick  must  the  floor  be  made 

WL 

to  hold  the  tank  full  of  water  ?   Allow  M  =  --    Compression  in  concrete 

10 

500  Ibs.  per  sq.  in.;  tension  in  steel  14,000  Ibs.  per  sq.  in. 

E, 


Find  also  the  area  of  the  steel  reinforcement. 

The  common  practice  in  concrete  building  construction  is  to  use  rein- 
forced-concrete  beams  rather  than  the  steel  beams.  In  this  event,  owing 
to  the  monolithic  character  of  the  work,  part  of  the  floor  slab  acts  as  the 
compression  flange  of  the  beam.  See  Chapter  XIV. 

164.  The  T  beams  have  a  span  of  20  ft.,  and  are  spaced  6  ft.  center  to 
center;  see  Fig.  59.  The  combined  live  and  dead  load  is  250  Ibs.  per  sq.  ft. 

Assuming  M  —  -  ;   •=*  =  15;  fc  =  500;  /,  =  14,000;  find  the  thickness 

IO         Ac 

of  the  slab. 


6  0-  -  ff< 


6  0- 


FIG.  59. 


FIG.  60. 


165.  Take  Problem  164;  assume  the  slab  5  ins.  thick  and  also  thatd  = 
1 6  ins.  for  the  T  beam;  find  the  width  of  slab  required  for  the  flange  of  the 
T  beam  and  the  area  of  the  reinforcing  metal. 

NOTE.  —  It  is  usual  to  limit  the  width  of  the  slab  assumed  as  flange  of  the 
T  beam.  One  specification  puts  this  limit  at  eight  times  the  slab  thickness 
but  not  exceeding  one-third  the  T-beam  span. 

Figure  60  shows  a  plan  of  a  section  of  a  floor.  The  following  data  apply 
to  all  these  problems. 

fc  =  55°  Ibs.,  /,  =  14,000 Ibs.,  —r  =  15. 

Live  load  150  Ibs.  per  sq.  ft.  of  floor,  dead  load  60  Ibs.  per  sq.  ft.  of  floor. 

166.  Design  the  slab,  determining  the  thickness  and  area  of  metal.    Take 


10 


408  GRAPHICS  AND  STRUCTURAL  DESIGN 

167.  Assume  the  slab  5!  ins.  thick,  and  design  the  T  beam.    Assume  d  = 
19  ins.    Find  width  of  flange  and  area  of  steel. 

168.  Design  the  girder  G,  assuming  that  it  carries  a  T  beam  B,  on  each 
side  of  it,  entering  it  centrally,  and  that  it  takes  only  85  per  cent  of  the  live 
load,  i.e.,  150  Ibs.  per  sq.  ft.,  but  entire  dead  load  of  75  Ibs.  per  sq.  ft.    As- 
sume the  depth  of  the  girder  as  d  =  25  ins.     It  must  be  at  least  deep  enough 
to  allow  its  reinforcing  bars  to  pass  under  the  bars  in  the  T  beams.     Find 
the  area  of  the  reinforcement  and  the  width  B  of  the  upper  flange  of  the  beam. 

Assume  the  bending  as  four-fifths  that  due  to  a  freely  supported  beam 

centrally  loaded,  i.e.,  M=  -- 

169.  A  beam  spans  20  ft.  and  carries  2500  Ibs.  per  ft.    Assume  that  d  =  30 
ins.  and  find  the  width  and  reinforcing  area.    Take/,  =  12,000  Ibs.,  fc  = 

iu     E*  ™     WL 

500  Ibs.,-5  =  15  and  M=  —  — 

-C«c  o 

170.  Take  the  data  given  and  found  for  Problem  169,  and  find  the  effect 
on/,  and/c  of  placing  the  steel  2  ins.  above  the  intended  position,  making 
d  =  28  ins.  instead  of  30  ins. 

171.  A  slab  spans  10  ft.  6  ins.  and  carries  a  load  of  485  Ibs.  per  sq.  ft. 

WL  E 

Assume  M  =  -  and  allow/,  =  16,000  Ibs.,  fe  =  600  Ibs.  and  —  -  —  15  Ibs. 
12  h,c 

Find  the  thickness  of  the  slab  and  the  spacing  of  f-in.  diameter  round  rods. 
Slabs  should  always  be  reinforced  in  both  directions,  the  metal  at  right 
angles  to  the  main  reinforcement  being  used  to  prevent  the  slab  from 
cracking.  When  the  slabs  are  designed  square  and  reinforced  in  both 
directions  the  maximum  moment  may  be  assumed  as 


20 

172.  A  square  slab  reinforced  in  both  directions  has  a  side  of  16  ft.  and 
carries  a  combined  live  and  dead  load  of  220  Ibs.  per  sq.  ft.     Find  thickness 
of  slab  and  area  of  reinforcement. 

173.  When  reinforced-concrete  beams  are  designed  assuming  that  the 

WL 

bending  at  the  center  is  -  ,  what  is  the  bending  at  the  supports?     Where 
10 

in  the  span  is  the  point  of  zero  bending  and  what  should  be  the  minimum 
reinforcing  over  the  supports  expressed  as  a  percentage  of  the  reinforcing 
at  the  center? 

In  the  problems  thus  far  considered  three  points  have  not  been  taken  into 
account.  The  horizontal  shear,  the  vertical  shear  and  diagonal  tension. 
See  Chapter  XIV. 


PROBLEMS 


409 


174.  Derive  the  following  formula  for  the  spacing  of  stirrups, 

o  0.85  XdxR 

~  F-(o.8s  Xvc  Xb  X<*)* 

175.  Derive  the  following  formula  for  the  length  of  rod  required  on  ac- 
count of  bonding, 


2  U 


176.  Derive  the  following  formula  for  the  length  of  bar  required  to 
provide  flange  strength, 


/  = 


177.  Take  Problem  158  and  determine  the  vertical  reinforcements,  allow- 
ing a  shearing  fiber  stress  of  12,000  Ibs.  per  sq.  in.  Neglect  the  shearing 
strength  of  the  concrete.  Try  f-in.  rounds  bent  into  U  stirrups.  Would 
stirrups  be  needed  if  50  Ibs.  per  sq.  in.  was  allowed  in  shear  in  concrete? 
What  about  reinforcement  at  the  ends? 

178  to  181.  Design  a  light  highway  bridge  of  reinforced  concrete.  Span 
32  ft.  o  ins.,  fill  15  ins.,  roadway  16  ft.  with  a  4-ft.  wall  on  each  side. 
Assume  8  =  5^  ins.  See  Figs.  61  to  63. 

FIG.  61. 


C— '""""""V^iiyz:'^.-.^.!. L.r.'.ul..  ..      ..'.      '.~~~^    t — — *.•    fl  X 

L  I      i          '       ~1'    '    ' 

— 5'o---i>*l — 5'o^4>l4--^'o^-4>J  I  GMi 


1^^  —  J  1^^^ 

40Tons  on  8  Wtieels 

1  - 

-H  sV'H  —  Car  Width  8V         \ 

:                                         i    1    . 

1 

o 

O                  QiO 

FIG.  62. 


W=20  Tons 

FIG.  63. 


178.  Determine  thickness  of  slabs  under  roadway.    Assume  live  load 
500  Ibs.  per  sq.  ft.,  concrete  60  Ibs.  per  sq.  ft.,  fill  1 20  Ibs.  per  sq.  ft.    Assume 


410  GRAPHICS  AND  STRUCTURAL  DESIGN 

compressive  stress  in  concrete  650  Ibs.  per  sq.  in.,  tensile  stress  in  steel 
16,000  Ibs.  per  sq.  in.  Assume  that  the  effect  of  impact  is  50  per  cent  of 
live-load  stresses. 

M  =  —     and     ^  =  15. 

IO  Ec 

Using  |-in.  round  bars,  how  far  apart  (center  to  center)  should  they  be 

spaced?     Since  the  bending  has  been  taken  at  M  = ,  what  reinforce- 

10 

ments  should  be  placed  over  girders? 

179.  Design  girders  d  and  G6.    Assume  dead  load  of  concrete,  fill,  etc., 
as  900  Ibs.  per  lineal  ft.  of  32-ft.  span.    Live  load  125  Ibs.  per  sq.  ft.  of  walk 
assumed  as  5  ft.  o  ins.  wide. 

z  =  1.64  corresponding  to  fc  =  650  Ibs.     M  =  — — . 

o 

1 80.  Design  girders  G3  and  G4  on  the  following  basis.    Each  girder  to 
carry  one-half  the  car  load  or  the  front  wheel  of  the  roller  and  one-half  the 
weight  of  the  rear  wheels,  Figs.  62  and  63.    Assume  dead  load  1350  Ibs. 

per  ft.     Take  M  =  — — .     Add  35  per  cent  to  live-load  stresses  to  allow  for 

o 

impact.  How  many  i|-in.  0  bars  will  be  required?  Assume  d  =  34.5, 
/,  =  16,000  Ibs.  What  width  of  the  flange  will  be  stressed?  Make  stem 
14  ins.  wide. 

181.  Draw  diagram  of  maximum  shears  covering  dead  load  and  trolley 
load.    Assume  ^-in.  0  U  stirrups  and  find  the  spacing  at  the  abutment  and 
at  2,  4,  6  and  8  ft.  from  this  end,  allowing  60  Ibs.  per  sq.  in.  shear  on  the 
concrete  and  using  the  formula 

s==          0.85  XdxR 

V  -  (0.85  Xvc  Xb  Xd)' 

R  =  shearing  value  of  one  stirrup. 

182.  In  a  given  beam  d  =  30  ins.,  four  i^-in.  round  reinforcing  bars  run 
parallel  with  the  lower  flange  through  the  entire  span.    The  maximum 
reaction  is  44,000  Ibs.     What  is  the  bond  stress  per  square  inch  and  is  this 
satisfactory  if  the  specifications  allow  60  Ibs.  per  sq.  in.?    How  could  this 
be  provided  for? 

183.  In  a  beam  d  =  34  ins.,  three  i-in.  0  bars  run  parallel  with  the 
lower  flange  through  the  entire  span.     The  maximum  reaction  is  19,000  Ibs. 
What  is  the  bond  stress  per  square  inch  and  is  this  satisfactory  if  the  speci- 
fications allow  a  bond  stress  of  60  Ibs.  per  sq.  in.?    If  not  satisfactory,  how 
could  it  be  provided  for? 


PROBLEMS 


411 


REINFORCED-CONCRETE  COLUMNS 

184.  Allowing  600  Ibs.  per  sq.  in.  on  the  concrete,  what  load  can  be  placed 
on  a  column  n  ft.  high,  12  X  12  ins.  and  reinforced  by  four  £-in.  0  bars? 
When  a  column's  height  does  not  exceed  twelve  times  its  least  dimension 
the  influence  of  length  upon  its  buckling  can  be  neglected.     Assume  column 
dimensions  inside  bars  as  10  X  10  ins. 

185.  Load  420,000  Ibs.,  reinforcement  i  per  cent,  height  15  ft.     What 
size  square  columns  allowing  600  Ibs.  upon  concrete  will  be  required?    How 
many  i-in.  0  bars  will  be  required?    What  would  you  make  the  outside 
dimensions  of  the  column? 

1 86.  Take  the  truss  in  Fig.  64,  place  a  load  L  at  the  point  indicated, 
show  the  necessity  for  the  double  lacing.     Explain  why  this  is  usual  in  the 
central  panels  and  frequently  not  necessary  at  the  end  panels. 


FIG.  64. 


FIG.  65 


187.  With  Fig.  65,  by  means  of  force  and  equilibrium  polygons  find  the 
required  depth  of  a  concrete  foundation  8  ft.  square,  at  125  Ibs.  per  cu.  ft. 
so  that  the  moment  of  the  foundation  about  the  toe  A  shall  be. twice  that 
of  the  crane  weight  and  live  load  about  the  same  point.    At  the  bottom  of 
the  foundation  make  a  diagram  showing  the  distribution  of  pressure  on  the 
soil. 

1 88.  Take  the  data  given  and  found  for  Problem  187 ;  turn  the  crane  until 
the  load  lies  in  the  line  of  a  diagonal  of  the  foundation.     With  the  resultant 
pressure  upon  the  soil  as  found  in  Problem  187,  determine  the  maximum  and 
minimum  pressure  on  the  foundation  by  the  method  given  in  Chapter  XVI. 

189.  Take  Fig.  66,  and  by  means  of  a  force  and  an  equilibrium  polygon 
determine  the  maximum  load  the  locomotive  crane  can  lift  in  the  position 
given.     Boom  is  36  ft.  long,  and  weighs  2000  Ibs.     Gauge  of  track  is  4  ft.  8^ 
ins.     Counterweight  is  15,000  Ibs.  acting  at  center  line  of  track.     Engine, 
machinery,  etc.,  weigh  18,000  Ibs.,  and  act  12  ins.  to  left  of  center  line  of 
track.    Boiler  and  base  weigh  12,000  Ibs.,  and  act  7  ft.  6  ins.  to  left  of  center 
line  of  track.    Cab,  etc.,  weigh  16,000  Ibs.,  and  act  at  center  line  of  track. 


412 


GRAPHICS  AND  STRUCTURAL  DESIGN 


190.  Design  a  concrete  footing,  Fig.  67,  to  carry  200  tons  from  the  column 
through  a  shoe  4  ft.  square,  resting  on  reinforced  concrete.  The  soil  can 
carry  3  tons  per  sq.  ft.  What  spacing  of  f-in.  round  bars  will  be  required? 
Assume 


fe  =  55o;  ft  =  16,000  Ibs.; 


=  15. 


FIG.  66. 


FIG.  67. 


What  advantages  would  this  type  of  foundation  have  over  the  usual  con- 
crete pier?     Consult  Chapter  XV. 

191.  In  Fig.  68  assume  sag  60  per  cent  of  width,  coal  (bituminous)  50 
Ibs.  per  cu.  ft.  Area  of  surcharged  bin  A'  =  0.57  a2.  Assume  that  the 
load  on  the  bin  varies  from  the  sides  to  the  middle  as  the  intercepts  between 


h- 


FIG.  68. 


the  sides  of  a  triangle.  Show  by  means  of  force  and  equilibrium  polygons 
how  to  find  the  tension  in  the  bottom  of  the  bin  plate  and  also  in  the  sides 
at  the  point  of  suspension  from  the  side  girders. 

192.  A  stack,  Fig.  69,  has  outside  diameter  of  7  ft.  and  is  120  ft.  high. 
Assume  a  uniform  wind  pressure  of  30  Ibs,  per  sq.  ft.  acting  on  it.    The  steel 


PROBLEMS  413 

weighs  30,000  Ibs.  and  the  concrete  base  is  20  it.  square.  If  the  concrete 
weighs  125  Ibs.  per  cu.  ft.,  how  deep  must  the  foundation  be  if  the  resultant 
pressure  falls  within  the  kern  and  what  is  the  distribution  of  pressure  on 
the  soil? 

193.  The  y-ft.  stack  shown  in  Fig.  69  is  120  ft.  high,  is  made  of  TVin.  plate 
for  30  ft.  at  the  top,  and  each  succeeding  30  ft.  toward  the  foot  of  the  stack 
is  made  ^  in.  thicker.    What  is  the  extreme  fiber  stress  in  the  net  section, 
60  ft.  from  the  top,  assuming  20  per  cent  of  the  plate  cut  out  for  rivets? 
Calculate  the  pitch  of  rivets,  allowing  8000  Ibs.  in  shear  and  16,000  Ibs.  in 
bearing  and  assume  rivets  f  in.  in  diameter. 

194.  In  Fig.  69  investigate  the  fiber  stress  and  rivet  pitch  for  the  seam 
80  ft.  from  the  top. 

195.  A  stack  is  125  ft.  high.    Its  outside  diameter  is  7  ft.  6  ins.     If 
12  bolts  are  used  on  a  circle  whose  radius  is  65  ins.,  what  must  their 
diameter  be?    Allowable  tension  is  12,000  Ibs.  per  sq.  in.    Weight  of  stack 
is  30,000  Ibs. 


ANSWERS   TO  PROBLEMS 


5- 

0.79  in. 

45- 

6. 

2.01  ins. 

7- 

3.82  ins. 

8. 

3.86  ins. 

14- 

13-5. 

IS- 

3.90. 

46. 

16. 

1334. 

47- 

17. 

14.6. 

48. 

18. 

324.0. 

49. 

19. 

241.1. 

50. 

20. 

0.0069  ^4- 

SI- 

21. 

11,340.1. 

53- 

24. 

W  T 

R  =  W,M=—  and   M 

rx  =  w?L.    54. 

3 

3  U       55- 

27. 

R-W,  M       -—  » 

58. 

2                                6 

59- 

Wr         9   Wr3 

_-               KK3C            ^     VV  X~ 

MX  =  FT" 

60. 

2         3    £2 

61. 

28. 

TF 

R\  =    5     :Vr,-,OT    =    O.^78    L, 

63. 

30. 


32. 

34. 
35- 

36. 

37- 
38. 
39- 

40. 

43- 
44- 


Ri=  7000  Ibs., 

Afmax=  78,ooo  ft.  Ibs., 

Mu  =  75,ooo  ft.  Ibs. 
-Mmax  =  446,490  ft.  Ibs., 

MM  =  443,600  ft.  Ibs. 
i5-in.  I  at  45  Ibs. 
i5-in.  I  at  50  Ibs. 

-required    217.4  use  two  2o-in.  I 

beams  at  65  Ibs. 
i5-in.  I  at  42  Ibs. 
i5-in.  I  at  42  Ibs. 
2o-in.  I  at  65  Ibs. 

-  =  126.9.     2o-in.  I  at  75  Ibs. 

6 

21,000  Ibs. 

Load  1460  Ibs.,  deflection  0.161  in. 


66. 
67. 
68. 
69. 
70. 

71- 
72. 

73- 
74- 
75- 
76. 

77- 
78. 

79- 
80. 


Load  19,200  Ibs.,  fiber  stress  in  fixed 
beam  10,670  Ibs.  per  sq.  in.,  de- 
flection in  supported  beam  0.534 
in.,  deflection  in  fixed  beam  0.107 
ins. 

2 1  ins.  diam. 

15,500  Ibs. 

4.92  ins.  back  to  back,    r  =  3.10. 

rQ  =  1.91,  n  =  1.73- 

r0  =  i.oo,  n  =  2.53. 

r  =2.0,  r  =  2.5. 

6150,  7600  and  6550  Ibs. 

11,400,  10,600  and  n, 750  Ibs. 

203,000  Ibs. 

34,880  Ibs. 

64,400  Ibs. 

66,300  Ibs. 

C  =  16,740  Ibs.;  P  =  19,100  Ibs. 

One  i2-in.  channel  at  20^  Ibs.  hori- 
zontally and  on  top  of  one  20-in. 
I  beam  at  65  Ibs.  placed  verti- 
cally. Lateral  stress  5700  Ibs. 
per  sq.  in. 

59.4  ins. 

5760  Ibs. 

6400  Ibs. 

16  ft.  8  ins. 

4740  Ibs.;   A  =  0.486  in. 

1420  Ibs. 

10,940  Ibs. 

8125  Ibs. 

8450  Ibs.  compression. 

1 1, 660  Ibs.  compression. 

7  rivets. 

12  rivets. 

10  rivets. 

81,000  Ibs. 

5  7,600  Ibs. 


414 


ANSWERS  TO   PROBLEMS 


415 


81.  52,500  Ibs. 

83.  21.8  sq.  ins. 

84.  24.9  sq.  ins. 

85.  21.7  sq.  ins. 

86.  Top  flange  two  4  X  4  X  £-in.  angles, 

and  one  plate  22  X  iVm-> 
lower  flange  two  4X4  Xf-in. 
angles  and  one  plate  22  X  i^-in. 
angles. 

87.  10,990  in.  Ibs.,  and  916  Ibs. 

88.  13,400  in.  Ibs.  and  1117  Ibs. 

89.  The  twisting  moment  of  a  round 

shaft  may  be  1.2  times  that  on  a 
square  shaft  of  the  same  weight 
per  foot. 

90.  H.P.  -   *«L 

32,130 

91.  21.3°. 

92.  2TV  ins. 

93.  2\  ins. 

94.  GC  =  37,600    Ibs.    comp.,    AG  = 

33,600  Ibs.  tension. 

95.  GH  =  4300  Ibs.  comp.,  DH  =  35,460 

Ibs.  comp. 

96.  HI  =  4800  Ibs.  tension  and  AI  = 

28,800  Ibs.  tension. 

97.  FL  —  31,180  Ibs.  comp.  andZJf  = 

14,400  Ibs.  tension. 

98.  BF  =  154,660  Ibs.  comp.,  FA  = 

109,375  Ibs.  tension. 

99.  FG  =  110,470  Ibs.  tension,  GC  = 

187,500  Ibs.  compression. 

100.  GH  =  31, 250  Ibs.  comp.,  —  El  — 

66,280  Ibs.  comp. 

114.  AL  =  187,500  Ibs.  comp.,  KE  = 
166,700  Ibs.  tension  and  KL  = 
32,500  Ibs.  tension. 
129.   18  rivets. 


130.  3,  the  spacing  is  commonly  not 
allowed  to  exceed  4  to  6  ins. 

135.  4.6  ins. 

136.  4.8  ins. 

137.  4.4  ins. 

138.  2.28  ins. 

139.  2.8  ins. 

140.  2.93  ins. 

141.  4.28  ins. 

142.  14.67  ins. 

143.  9.6  ins. 

144.  0.27  in. 

145.  2  rows. 

146.  Two  plates  each  f  in.  thick. 

147.  2  rows. 

148.  6  rivets. 

149.  Check  by  method  of  coefficients. 

158.  b  =  13.5  ins.  and  A  =  i-34sq.  ins. 

159.  d  =  24  ins.,  6  =  9.5  ins.,  A   = 

1.14  sq.  ins. 

161.  d  =  2.2  ins.  and  A  =  0.166  sq.  in. 

162.  d  =  4.9  ins.  and  A  =  0.372  sq.  in. 

169.  b  =  20  ins.  and  A  =  4.8  sq.  ins. 

170.  fc  =  555  Ibs.  per  sq.  in.  and  /,  = 

1 2, 800  Ibs. 

171.  d  =  6.85  ins.     Spacing  =  6.77  ins. 

178.  d  =  4.65  ins.,  j-in.  0  bars  5-5-in. 

centers. 

179.  d  =  38.0  ins.,  five  —  i£-in.  0  bars. 

180.  eight  i^-in.  0  bars  or  nine  i£-in. 

0  bars. 

182.  93  Ibs.  per  sq.  in. 

183.  59  Ibs.  pef  sq.  in. 

184.  66,590  Ibs. 

185.  25  ins.  square  and  eight  i-in.  0 

bars. 

190.  |-in.  0  rods  spaced  4.75  ins.  center 
to  center. 


INDEX 


Algebraic  determination  of  stresses,  52. 
Angles  of  repose  and  weights  of  materials,  277. 
Arch  floors,  brick  and  tile,  320. 
Ash  bins,  pressures  on,  290. 

Base  for  reinforced-concrete  chimney,  270. 

Beams,  design  of,  76. 

Beams  of  uniform  section,  bending  moments,  deflections,  etc.,  7. 

Beams,  reinforced-concrete: 

approximate  formulae  for,  196. 

bent  bars,  217. 

bond  stresses,  214. 

design  of  a  T  beam,  216. 

diagonal  tension  in,  213. 

forms  of  reinforcements,  223. 

horizontal  shear  in,  209. 

parabolic  variation  of  stresses,  208. 

reinforcing  rods,  lengths  of,  215. 

stirrup  spacing  in,  211. 

theory  of  beams  with  double  reinforcement,  205. 
rectangular  beams,  194. 
slabs,  204. 
T  beams,  199. 

Beams  unsupported  laterally,  98. 
Bearing  and  shearing  value  of  rivets,  84. 
Bed  plates  for  plate  girders  for  railway  bridge,  159. 
Bending  due  to  moving  loads,  47. 
Bending  moments  on  beams,  77. 

Bending  moments,  deflections,  etc.,  for  beams  of  uniform  section,  7. 
Bending  moments  on  a  railway  girder  due  to  locomotive  and  train  load,  141. 
Bents,  112. 

Bent  bars  in  T  beams,  217. 
Bethlehem  rolled  sections,  18. 
Bins: 

coefficients  of  friction  between  materials,  296. 

design  of,  306. 

graphical  determination  of  forces  acting  on,  296. 

417 


41 8  INDEX 

Bins: 

pressure  on  grain  bins,  289. 
shallow  bins,  290. 

stresses  in  bins,  formulae  for,  290. 
Bond  stresses  between  concrete  and  steel,  214. 
Box  girder  for  electric  overhead  traveling  crane,  179. 
Bracing  for  steel  mill  buildings,  118. 
Bricks,  4. 
Bricks,  fire,  4. 
Brick  arch  floors,  319. 
Brick  chimney,  design  of,  242. 
Brick  floors,  316. 
Brick  walls,  323. 
Buildings,  foundations  for,  230. 
Bunkers,  suspension,  300. 

Cast  iron,  i. 

Cement,  191. 

Cement  concrete  floors,  314. 

Center  of  gravity,  graphical  determination  of,  30. 

Centers  of  gravity  of  angles,  15. 

channels,  16. 
Character  of  stresses,  43. 
Chimney,  brick: 

design  of  a,  242. 

kern  of  a  section,  238. 
Chimney,  reinforced-concrete:    • 

base,  270. 

design  of  a  reinforced-concrete,  259. 

reinforcements,  267. 
Chimney,  steel: 

design  of  a,  248. 

foundation,  253. 

foundation  bolts,  257. 

lining,  253. 

maximum  pressure  on  the  soil,  255. 

ring  seams,  251. 

thickness  of  shell,  248. 

vertical  seams,  251. 
Clay  tile  roofing,  341. 
Clearances  for  cranes,  336. 
Coal  bins,  pressures  on,  290. 
Column  formulae,  90. 
Columns  for  steel  mill  buildings,  126. 
Columns,  90. 
Columns,  theory  of  reinforced-concrete,  221. 


INDEX  419 


Columns,  timber,  4. 

Combined  stresses,  95. 

Compression  pieces,  design  of,  76. 

Concrete  (see  also  Reinforced  concrete). 

Concrete  blocks,  hollow,  325. 

Concrete  retaining  wall,  design  of,  279. 

Concrete  roofing,  341. 

Concrete  walls,  solid,  324. 

Concrete  wearing  surfaces,  321. 

Concurrent  and  nonconcurrent  forces,  21. 

Continuous  beams,  32. 

Conveyors,  girders  for,  101. 

Cooper's  E-6o  loading,  moment  table  for,  71. 

Coplanar  and  noncoplanar  forces,  21. 

Corrugated  bars,  224. 

Corrugated  steel  roofing,  337. 

Corrugated  steel  sides,  323. 

Couple,  21. 

Cranes: 

electric  overhead  traveling  crane,  174. 

top-braced  jib  crane,  168. 

underbraced  jib  crane,  162. 
Crane  clearances,  336. 
Crane  frames  (see  Cranes). 

Deflection  of  beams,  graphical  determination  of,  30. 
Deflection,  bending  moments,  etc.,  7. 
Design  of  beams,  77. 

compression  pieces,  76. 
frame  for  top-braced  jib  crane,  168. 
frame  for  underbraced  jib  crane,  162. 
railway  girder,  141. 
roofs,  337. 

roof  truss  members,  123. 
steel  mill  building,  No.  i,  118. 
steel  mill  building,  No.  2,  130. 
tension  pieces,  76. 
Determination  of  stresses,  algebraic,  52. 

method  of  coefficients,  55, 
method  of  moments,  54. 

Diagonal  tension  in  reinforced  concrete  beams,  212. 
Diagram  of  maximum  live-load  shears,  49. 
Dimension  of  angles,  14. 

Bethlehem  beams,  19. 
channels,  16. 
edged  plates,  13. 


420  INDEX 

Dimension  of  angles  (continued). 

girder  beams,  19. 

I  beams,  17. 

H  columns,  18. 

sheared  plates,  12. 
Direction  of  a  force,  20. 

Elastic  limit  of  concrete,  192. 
Electric  overhead  traveling  crane: 
design  of  box  girder,  175. 
flange  areas,  176. 
rivets,  180. 

girder  with  channel  flanges,  182. 
girder  with  horizontal  stiffening  girder,  184. 
stiff  eners,  182. 
specifications  for,  174. 
Equilibrant,  21. 
Equilibrium,  21. 

polygon,  23. 
Expanded  metal,  222. 

Fiber  stresses  in  underbraced  jib  cranes,  164. 

working,  3. 
Flange  areas  for  girders  of  E.O.T.  cranes,  176. 

of  plate  girders,  146. 

Flange  plates  of  plate  girders  for  railway  bridges,  lengths  of,  148. 
Flange  rivets  in  girder  for  E.O.T.  crane,  180. 
Flange  rivets  in  plate  girder  for  railway  bridge,  149. 
Flat  plates,  strength  of,  9. 
Floors,  ground: 

brick,  316. 

cement  concrete,  314. 

wooden,  317. 

wooden  block,  316. 
Floors,  upper: 

brick  arch,  319. 

concrete  wearing  surfaces,  321. 

hollow  tile  arch,  320. 

iron  floors,  322. 

reinforced-concrete,  321. 

steel,  319. 

steps,  322. 

wooden,  318. 

Floor-beam  reaction,  maximum,  69. 
Force,  20. 
Force  and  equilibrium  polygons,  uses  of,  28. 


INDEX  421 

Force  polygon,  22. 

Force  triangle,  21. 

Forces  acting  in  bins,  graphical  determination  of,  296. 

Formulae  for  piles,  237. 

for  stresses  in  bins,  291. 
Foundation  bolts  for  chimneys,  257. 
Foundations  for  buildings,  230. 

for  machinery,  225. 

piles,  235. 

pile  formulae,  237. 

pressure  on  soil,  allowable,  228. 
Foundations  for  chimneys,  253. 
Framing,  standard,  80. 
Friction,  coefficient  of,  between  materials,  296. 

Girders  for  conveyors,  101. 

Girders  for  E.O.T.  crane,  specifications  for,  174. 

with  channel  flanges  for  E.O.T.  cranes,  182. 

with  horizontal  stiffening  girders  for  E.O.T.  cranes,  184. 
'Glass,  337. 

Grain  bins,  pressures  on,  289. 
Graphic  moments,  26. 
Graphics,  20. 

statics,  20. 
Graphical  determination  of  forces  acting  in  bins,  296. 

of  stirrup  spacing  in  reinforced-concrete  beams,  211. 
Gravel,  191. 

Gravel  and  slag  roofing,  341. 
Ground  floors,  317. 

Hollow  tile  arch  floors,  320. 

Horizontal  shear  in  reinforced-concrete  beams,  209. 

stiffening  girders  for  E.O.T.  crane  girders,  184. 

Impact,  143. 

Inertias  of  geometrical  sections,  10. 

angles,  14. 

channels,  ^16. 

H  columns,  18. 

I  beams,  17. 
Influence  diagrams,  61. 
Iron,  cast,  i. 
Iron  floors,  322. 

Jib  for  underbraced  crane,  selection  of,  164. 
Jib  crane,  underbraced,  162. 
top-braced,  168. 


^T 


422  INDEX 

Kahn  trussed  bar,  224. 
Kern  of  a  section,  238. 

Lateral  bracing  for  plate-girder  railway  bridge,  153. 

Line  of  action,  20. 

Lining,  chimney,  253. 

Live-load  shears,  diagram  of  maximum,  49. 

stresses,  56. 
Long  beams  unsupported  laterally,  98. 

Machinery,  foundations  for,  225. 

Magnitude  of  a  force,  20. 

Masonry  retaining  walls,  design  of,  279. 

Mast  of  a  top-braced  jib  crane,  173. 

Mast  of  an  under-braced  jib  crane,  166. 

Materials,  i. 

Maximum  bending  due  to  moving  loads,  47. 

floor-beam  reaction,  69. 

live-load  shears,  diagram  of,  49. 

moment  due  to  moving  loads,  61. 

shear  due  to  moving  loads,  64. 
Members  of  frame  of  top-braced  jib  crane,  166. 

of  underbraced  jib  crane,  170. 
Metals,  physical  properties  of,  2. 
Method  of  coefficients  for  the  determination  of  stresses,  55. 

moments,  determination  of  stresses  by,  54. 
Mixing  concrete,  191. 
Moment  of  a  couple,  21. 

table  for  Cooper's  E-6o  loading,  71. 
Moving  loads  carried  under  trusses,  37. 

maximum  bending  due  to,  47. 

Physical  properties  of  concrete,  191. 

metals,  2. 
Pile  formulae,  237. 
Piles,  235. 
Plate  girder  for  railway  bridge: 

bed  plates,  159. 

flange  rivets,  149. 

flange  plates,  lengths  of,  148. 

rivet  spacing  in  flange  angles,  151. 

stiffening  angles,  148. 

web  splice,  154. 

wind  bracing,  153. 
Plate  girders: 

dead-load  shear,  146. 

maximum  end  shear,  144. 


INDEX  423 


Plate-girder  railway  bridge,  141. 
Plates,  strength  of  flat,  9. 

dimensions  of  edged,  9. 

of  sheared,  9. 

Point  of  application  of  a  force,  20. 
Pressures  on  retaining  walls,  273. 
soils,  allowable,  228. 
soil  under  chimney,  maximum,  255. 
Properties  of  concrete,  physical,  192. 
metals,  2. 
sections,  10. 
Properties  of  timber,  3. 

table  of,  5. 
Purlins,  126. 


Railway  girder,  design  of,  141. 

girders,  impact  allowance  for,  143. 
Ransome  twisted  bars,  224. 
Rectangular  beams,  theory  of,  194. 
Reinforcing  frame,  224. 

rods,  lengths  of,  215. 
Reinforced  concrete: 

approximate  formulae  for  rectangular  beams,  196. 

bond  stresses,  214. 

cement  for,  191. 

columns,  221. 

design  of  T  beams,  216. 

diagonal  tension,  213. 

elastic  limit,  192. 

forms  of  reinforcements,  223. 

gravel,  191. 

horizontal  shear  in  beams,  209. 

mixing,  191. 

parabolic  variation  of  stress,  208. 

physical  properties,  191. 

reinforcing  rods,  lengths  of,  215. 

sand,  191. 

stone,  191. 

stirrup  spacing  in  beams,  211. 

theory  of  beams  with  double  reinforcement,  205. 
rectangular  beams,  194. 
slabs,  204. 
T  beams,  199. 
Reinforced-concrete  chimney,  259. 

retaining  walls,  280. 


424  INDEX 

Reinforced-concrete  walls,  325. 
Reinforcements,  forms  of,  222. 
Resistances  of  sections,  10. 
Resultant,  21. 
Retaining  walls: 

design  of  a  masonry,  279. 

reinforced-concrete,  280. 

distribution  of  pressure  on,  275. 

graphical  determination  of  the  pressures  on,  274. 

pressures  on,  273. 

weights  and  angles  of  repose  of  materials  in  fills,  277. 
Ring  seams  of  steel  chimneys,  251. 
Rivet  spacing  in  flange  angles,  151. 
Rivet  values  in  shear  and  bearing,  table  of,  84. 
Riveting,  81. 
Riveting,  rules  for,  88. 
Roofs,  types  of,  121. 
Roof  coverings: 

clay  tile,  341. 

concrete,  341. 

corrugated  steel,  337. 

slag  or  gravel,  341. 

slate,  340. 

Sand,  191. 

Sections,  inertias,  resistances,  etc.,  10 

properties  of,  10. 
Shallow  bins,  pressures  on,  290. 
Shear  and  bearing  values  of  rivets,  84. 
Shear  in  reinforced-concrete  beams,  horizontal,  209. 
Shell  thickness  of  steel  chimney,  248. 
Shop  floors  (see  Floors). 
Slag  or  gravel  roofing,  341. 
Slate  roofing,  340. 
Soil,  allowable  pressure  on,  228. 
Specifications  for  girders  of  an  E.O.T.  crane,  174, 
Splice  for  plate-girder  of  railway  bridge,  web,  154. 
Standard  framing,  79. 
Steel  castings,  2. 
Steel  floors,  319. 
Steel  mill  buildings: 

design  of  Problem  i,  118. 

design  of  Problem  2,  130. 
Steps,  322. 
Stiffeners  for  girders  for  E.O.T.  crane,  182. 


INDEX  425 


Stifteners  for  plate  girders  of  railway  bridge,  148. 
Stiffening  steel  mill  buildings,  118. 
Stirrup  spacing  in  reinforced-concrete  beams,  211. 
Stone,  191. 

walls,  324. 

Strength  of  flat  plates,  9. 
Stresses: 

algebraic  determination  of,  52. 

character  of,  43. 
Stresses  in  concrete,  unit  working,  192. 

structures,  35. 
Structural  material,  n. 
Structures,  stresses  in,  35. 

T  beams: 

bent  bars,  217. 

design  of  a,  216. 

web  reinforcements,  218. 
Tension  pieces,  design  of,  76. 
Tension  or  compression  in  a  member,  36. 
Thatcher  bar,  224. 
Theory  of  columns,  221. 
Tile  arch  floors,  320. 
Timber  columns,  4. 
Timber,  properties  of,  3. 

table  of  properties  of,  5. 
Top-braced  jib  crane: 

design  of  frame,  168. 

mast,  173. 

selection  of  members,  170. 
Towers,  112. 

Traveling  crane  (see  Electric  overhead  traveling  crane),  174. 
Triangular  mesh  steel  reinforcement,  224. 
Truss,  determination  of  the  stresses  using  moment  table,  74. 
Trusses,  no. 
Trusses  carrying  moving  loads  under  them,  37. 

wind  loads  on,  39. 
Twisted  bars,  Ransome,  224. 

Underbraced  jib  crane: 

design  of  frame,  162. 

fiber  stresses  in,  164. 

mast,  1 66. 

members  of  frame,  166. 

selection  of  jib,  164. 
Unit  working  stresses  for  concrete,  192. 


426  INDEX 

Vertical  seams  of  a  steel  chimney,  251. 
shear  in  beams,  77. 

Walls: 

brick,  323. 

corrugated  steel,  323. 

glass,  337. 

hollow  concrete  blocks,  325. 

reinforced-concrete,  325. 

solid  concrete,  324. 

stone,  324. 

wooden,  323. 
Wearing  surfaces,  321. 
Web  reinforcements  in  T  beams,  209. 
Web  splice  for  plate  girders  of  railway  bridge,  154. 
Weights  and  angles  of  repose  of  materials,  277. 
Wind  bracing  for  girders  carrying  conveyors,  107. 
plate-girder  railway  bridge,  153. 
steel  mill  building,  128. 
Wind  load  on  trusses,  39. 
Wooden  block  floors,  316. 
Wooden  floors,  318. 

walls  or  sides,  323. 
Working  fiber  stresses,  3. 


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MISCELLANEOUS. 

BURT — Railway  Station  Service 12mo,  *2  00 

CHAPIN — How  to  Enamel 12mo,  *1  00 

FERREL — Popular  Treatise  on  the  Winds 8vo,     4  00 

FITZGERALD — Boston  Machinist 18mo,     1  00 

FRITZ — Autobiography  of  John 8vo,  *2  00 

GANNETT — Statistical  Abstract  of  the  World 24mo,     0  75 

GREEN — Elementary  Hebrew  Grammar 12mo,      1   25 

HAINES — American  Railway  Management 12mo,     2  50 

HANAUSEK — The  Microscopy  of  Technical  Products.      (WiNTON.) 8vo,     5  00 

JACOBS — Betterment  Briefs.     A  Collection  of  Published  Papers  on  Organ- 
ized Industrial  Efficiency 8vo,     3  50 

METCALFE — Cost-  of  Manufactures,  and  the  Administration  of  Workshops. 

8vo,     5  00 

PARKHURST — Applied  Methods  of  Scientific  Management 8vo,  *2  00 

PUTNAM — Nautical  Charts ! 8vo,     2  00 

RICKETTS — History  of  Rensselaer  Polytechnic  Institute,  1824-1894. 

Small  8vo,     3  00 
ROTCH  and  PALMER — Charts  of  the  Atmosphere  for  Aeronauts  and  Aviators. 

Oblong  4to,  *2  00 

ROTHERHAM — Emphasised  New  Testament Large  8vo,     2  00 

RUST — Ex- Meridian  Altitude,  Azimuth  and  Star-finding  Tables 8vo,     5  00 

STANDAGE— Decoration  of  Wood,  Glass,  Metal,  etc 12mo,     2  00 

WESTERMAIER — Compendium  of  General  Botany.     (SCHNEIDER.) 8vo,     2  00 

WINSLOW — Elements  of  Applied  Microscopy 12mo,     1  50 


22 


u  a 


THIS  BOOK  IS  DUE  ON  THE  LAST  DATE 
STAMPED  BELOW 

AN  INITIAL  FINE  OF  25  CENTS 

WILL  BE  ASSESSED  FOR  FAILURE  TO  RETURN 
THIS  BOOK  ON  THE  DATE  DUE.  THE  PENALTY 
WILL  INCREASE  TO  SO  CENTS  ON  THE  FOURTH 
DAY  AND  TO  $1.OO  ON  THE  SEVENTH  DAY 
OVERDUE. 


JUN    28  1934 
IN  29  1934 


230c'52Mtj 


REC 


RJ 


LD  21-100m-7,'33 


GENERAL  LIBRARY  -  U.C.  BERKELEY 


271051 

-  —  ~ 


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UNIVERSITY  OF  CALIFORNIA  LIBRARY 


